first step in solving projectile motion problems

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Classical Mechanics

How to Solve a Projectile Motion Problem

Last Updated: October 6, 2017

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Projectile motion is often one of the most difficult topics to understand in physics classes. Most of the time, there is not a direct way to get the answer; you need to solve for a few other variables to get the answer you are looking for. This means in order to find the distance an object traveled, you might first have to find the time it took or the initial velocity first. Just follow these steps and you should be able to fly through projectile motion problems!

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1 thought on “Steps To Solve Projectile Motion Questions”

i love the way you give your steps but it does not come with diagram you people should try and give diagram in your steps to enable fast learning

Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

x = v i x •t + 0.5*a x *t 2

v f x = v i x + a x •t

v f x 2 = v i x 2 + 2*a x •x

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s . Once this cancellation of ax terms is performed, the only equation of usefulness is:

x = v i x •t

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

y = v iy •t + 0.5*a y *t 2

v fy = v iy + a y •t

v fy 2 = v iy 2 + 2*a y •y

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), v iy = 0 m/s. Thus, any term with v iy in it will cancel out of the equation.

The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, v ix , v iy , a x , a y , and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = v iy •t +0.5•a y •t 2 ) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

By taking the square root of both sides of the equation, the time of flight can then be determined .

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information , v ix = 2.4 m/s and a x = 0 m/s/s. The first horizontal equation (x = v ix •t + 0.5•a x •t 2 ) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold . Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

Check Your Understanding

Use y = v iy • t + 0.5 • a y • t 2 to solve for time; the time of flight is 2.12 seconds.

Now use x = v ix • t + 0.5 • a x • t 2 to solve for v ix

Note that a x is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the v ix can be found to be 16.5 m/s.

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Section Learning Objectives

By the end of this section, you will be able to do the following:

Teacher Support

The learning objectives in this section will help your students master the following standards:

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:

(C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

Section Key Terms

Properties of projectile motion.

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile , and its path is called its trajectory . Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.

[BL] [OL] Review addition of vectors graphically and analytically.

[BL] [OL] [AL] Explain the term projectile motion. Ask students to guess what the motion of a projectile might depend on? Is the initial velocity important? Is the angle important? How will these things affect its height and the distance it covers? Introduce the concept of air resistance. Review kinematic equations.

The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are independent , meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.

Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.

A cannonball is sitting on a flat surface. Dotted lines show the projectile motion of a cannon ball. Dotted line A shows the ball dropping vertically. Dotted line B shows a curved path. Dotted line C shows the ball moving horizontally.

We’ll call the horizontal axis the x -axis and the vertical axis the y -axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y , as illustrated in Figure 5.28 .

A boy is kicking a soccer ball. A curved line shows the ball's motion. A right triangle is drawn with vectors using an x-y axis. The hypotenuse is labeled d, the vertical leg is labeled y and the horizontal leg is labeled x.

As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x - and y -axes. The components of acceleration are then very simple a y = – g = –9.80 m/s 2 . Note that this definition defines the upwards direction as positive. Because gravity is vertical, a x = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1 .

Where x is position, x 0 is initial position, v is velocity, v avg is average velocity, t is time and a is acceleration.

Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:

Four related diagrams of projectile motion are show, a, b, c, and d. Diagram a shows two-dimensional curved motion, vx and vy, along an x-y axis and includes the formulas V y equals negative V oy and angle equals negative angle o. Diagram b shows horizontal motion, vx. Diagram c shows upward and downward vertical velocity, vy. And diagram d shows the combination of vx, vy, and v, forming a right triangle.

Teacher Demonstration

Demonstrate the path of a projectile by doing a simple demonstration. Toss a dark beanbag in front of a white board so that students can get a good look at the projectile path. Vary the toss angles, so different paths can be displayed. This demonstration could be extended by using digital photography. Draw a reference grid on the whiteboard, then toss the bag at different angles while taking a video. Replay this in slow motion to observe and compare the altitudes and trajectories.

Tips For Success

For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for x x and y y . Usually, it is simplest to set the initial position of the object so that x 0 = 0 x 0 = 0 and y 0 = 0 y 0 = 0 .

Watch Physics

Projectile at an angle.

This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle.

Worked Example

A fireworks projectile explodes high and away.

During a fireworks display like the one illustrated in Figure 5.30 , a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

The diagram shows the trajectory of a fireworks shell. An x-axis is labeled x, a y-axis is labeled h, and a vector, labeled V o, points upward from the axis origin. An angle is formed by the vector and the x-axis with a measure of seventy-five degrees.

The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and a y = g a y = g . We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the maximum height .

By height we mean the altitude or vertical position y y above the starting point. The highest point in any trajectory, the maximum height, is reached when v y = 0 v y = 0 ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v y when the firework reaches its maximum height, we use the following equation to find y y

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y y gives

Now we must find v 0 y v 0 y , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ = 75 ∘ θ = 75 ∘ is the initial angle. Thus,

Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t . Because y 0 y 0 is zero, this equation reduces to

Note that the final vertical velocity, v y v y , at the highest point is zero. Therefore,

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 g t 2 y = y 0 + v 0 y t − 1 2 g t 2 , and solving the quadratic equation for t t .

Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t , where x 0 x 0 is equal to zero

where v x v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . Now,

The time t t for both motions is the same, and so x x is

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the forces of the explosion.

[BL] [OL] [AL] Talk about the sample problem. Discuss the variables or unknowns in each part of the problem Ask students which kinematic equations may be best suited to solve the different parts of the problem.

The expression we found for y y while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height y = h y = h ; then,

This equation defines the maximum height of a projectile . The maximum height depends only on the vertical component of the initial velocity.

Calculating Projectile Motion: Hot Rock Projectile

Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31 , with a speed of 25.0 m / s 25.0 m / s and at an angle 3 5 ° 3 5 ° above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

A volcano is shown with labels illustrating the projectile motion of rock shooting out of the volcano. Vector v o equals twenty-five meters per second and points from the volcano opening upward. Vector v curves downward to the ground and forms an angle of thirty-five degrees with a horizontal plane at the height of the volcano opening. The distance between the horizontal plane and the ground is labeled as negative twenty meters.

Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 to be zero, then the final position is y = − 20.0 m . y = − 20.0 m . Now the initial vertical velocity is the vertical component of the initial velocity, found from

Substituting known values yields

Rearranging terms gives a quadratic equation in t t

This expression is a quadratic equation of the form a t 2 + b t + c = 0 a t 2 + b t + c = 0 , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m / s 14.3 m / s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Practice Problems

The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32 . Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.

Two trajectory diagrams are shown. Diagram a shows three trajectories that curve upward then downward. They are labeled as follows: thirty meters per second and ninety-eight point one meters; forty meters per second and one hundred sixty-three meters; fifty meters per second and two hundred fifty-five meters. Diagram b shows three trajectories that curve upward then downward. They are labeled as follows: fifteen degrees and one hundred twenty-eight meters; forty-five degrees and two hundred fifty-five meters; seventy-five degrees and one hundred twenty-eight meters. The diagram also includes the formula v o equals fifty meters per second.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 , the greater the range, as shown in the figure above. The initial angle θ 0 θ 0 also has a dramatic effect on the range. When air resistance is negligible, the range R R of a projectile on level ground is

where v 0 v 0 is the initial speed and θ 0 θ 0 is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

Virtual Physics

Projectile motion.

In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target.

Check Your Understanding

What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance?

Use the Check Your Understanding questions to assess whether students achieve the learning objectives for this section. If students are struggling with a specific objective, the Check Your Understanding will help identify which objective is causing the problem and direct students to the relevant content.

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Two-Dimensional Kinematics

Projectile motion, learning objectives.

By the end of this section, you will be able to:

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 1 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A to represent a vector with components A x and A y . If we continued this format, we would call displacement s with components s x and s y . However, to simplify the notation, we will simply represent the component vectors as x and y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x – and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = – g = –9.80 m/s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x =0. Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant a )

A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.

Figure 1. The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is s, and it makes an angle θ with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ and A y = A sin θ are used. The magnitude of the components of displacement s along these axes are x and y. The magnitudes of the components of the velocity v are V x = V cos θ and V y = v sin θ where v is the magnitude of the velocity and θ is its direction, as shown in 2. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s and velocity v. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing [latex]A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}\\[/latex] and θ = tan −1 ( A y / A x ) in the following form, where θ is the direction of the displacement s and θ v is the direction of the velocity v :

Total displacement and velocity

In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east.

Figure 2. (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because ax=0 and vx is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x – and y -motions are recombined to give the total velocity at any given point on the trajectory.

Example 1. A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = – g . We can then define x 0 and y 0 to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y =0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y :

The x y graph shows the trajectory of fireworks shell. The initial velocity of the shell v zero is at angle theta zero equal to seventy five degrees with the horizontal x axis. The fuse is set to explode the shell at the highest point of the trajectory which is at a height h equal to two hundred thirty three meters and at a horizontal distance x equal to one hundred twenty five meters from the origin.

Figure 3. The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 and v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , the component of the initial velocity in the y -direction. It is given by v 0y = v 0 sin θ , where v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º is the initial angle. Thus,

v Oy = v 0 sin θ 0 = (70.0 m/s)(sin 75º) = 67.6 m/s

[latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex] ,

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. Because y 0 is zero, this equation reduces to simply

[latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex].

Note that the final vertical velocity, v y , at the highest point is zero. Thus,

[latex]\begin{array}{lll}t& =& \frac{2y}{\left({v}_{0y}+{v}_{y}\right)}=\frac{2\left(\text{233 m}\right)}{\left(\text{67.6 m/s}\right)}\\ & =& 6.90\text{ s}\end{array}\\[/latex].

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex], and solving the quadratic equation for t .)

Solution for (c)

Because air resistance is negligible, a x =0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t , where x 0 is equal to zero:

x = v x t ,

where v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 Now,

v x = v 0 cos θ 0 = (70.0 m/s)(cos 75º) = 18.1 m/s

The time t for both motions is the same, and so x is

x = (18.1 m/s)(6.90 s) = 125 m.

Discussion for (c)

In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h ; then,

[latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

Example 2. calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º above the horizontal, as shown in Figure 4. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

The trajectory of a rock ejected from a volcano is shown. The initial velocity of rock v zero is equal to twenty five meters per second and it makes an angle of thirty five degrees with the horizontal x axis. The figure shows rock falling down a height of twenty meters below the volcano level. The velocity at this point is v which makes an angle of theta with horizontal x axis. The direction of v is south east.

Figure 4. The trajectory of a rock ejected from the Kilauea volcano.

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θ v at the final time t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex].

If we take the initial position y 0 to be zero, then the final position is y = −20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from v Oy = v 0 sin θ 0 = (25.0 m/s)(sin 35.0º) = 14.3 m/s. Substituting known values yields

Rearranging terms gives a quadratic equation in t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 , where the constants are a = 4.90 , b = –14.3 , and c = –20.0. Its solutions are given by the quadratic formula:

[latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]

This equation yields two solutions: t = 3.96 and t = –1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s or -1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x and v y and combine them to find the total velocity v and the angle θ 0 it makes with the horizontal. Of course, v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

v x = v 0 cos θ 0 = (25.0 m/s)(cos 35º) = 20.5 m/s

The final vertical velocity is given by the following equation:

[latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex],

where v 0y was found in part (a) to be 14.3 m/s. Thus,

To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:

[latex]v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{({20.5}\text{ m/s})^{2}+{({-24.5}\text{ m/s})^{2}}}\\[/latex]

which gives

The direction θ v is found from the equation:

The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 4.)

Part a of the figure shows three different trajectories of projectiles on level ground. In each case the projectiles makes an angle of forty five degrees with the horizontal axis. The first projectile of initial velocity thirty meters per second travels a horizontal distance of R equal to ninety one point eight meters. The second projectile of initial velocity forty meters per second travels a horizontal distance of R equal to one hundred sixty three meters. The third projectile of initial velocity fifty meters per second travels a horizontal distance of R equal to two hundred fifty five meters.

Figure 5. Trajectories of projectiles on level ground. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0 on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 , the greater the range, as shown in Figure 5(a). The initial angle θ 0 also has a dramatic effect on the range, as illustrated in Figure 5(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45º. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º. Interestingly, for every initial angle except 45º, there are two angles that give the same range—the sum of those angles is 90º. The range also depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by

[latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex],

where v 0 is the initial speed and θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 6.) If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

A figure of the Earth is shown and on top of it a very high tower is placed. A projectile satellite is launched from this very high tower with initial velocity of v zero in the horizontal direction. Several trajectories are shown with increasing range. A circular trajectory is shown indicating the satellite achieved its orbit and it is revolving around the Earth.

Figure 6. Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

PhET Explorations: Projectile Motion

Click to run the simulation.

Section Summary

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x and y , and the components of the velocity v are given by v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction.

2. Analyze the motion of the projectile in the horizontal direction using the following equations:

Horizontal Motion ( a x = 0)

3. Analyze the motion of the projectile in the vertical direction using the following equations:

Vertical Motion (assuming positive is up a y = -g = -9.8 m/s 2 )

4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:

Conceptual Questions

1. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at t = 0?

2. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

3. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

Problems & Exercises

1. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

2. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

4. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

5. An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

7. Verify the ranges for the projectiles in Figure 5 (a) for θ = 45º and the given initial velocities.

8. Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles.

9. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37 × 10 3 . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

10. An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

11. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

12. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

13. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

14. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25º relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

15. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

17. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

18. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.

19. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

21. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)

22. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

23. A football player punts the ball at a 45º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

24. Prove that the trajectory of a projectile is parabolic, having the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex]. To obtain this expression, solve the equation [latex]x={v}_{0x}t\\[/latex] for t and substitute it into the expression for [latex]y={v}_{0y}t-\left(1/2\right){\text{gt}}^{2}\\[/latex]. (These equations describe the x and y positions of a projectile that starts at the origin.) You should obtain an equation of the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex] where a and b are constants.

25. Derive [latex]R=\frac{{{v}_{0}}^{2}\text{\sin}{2\theta }_{0}}{g}\\[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t into the expression for x – x 0 , noting that R = x – x 0 .

26. Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

Selected Solutions to Problems & Exercises

1. x = 1.30 m × 10 2 , y = 30.9 m

3. (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2º below horizontal

5. (a) 18.4º (b) The arrow will go over the branch.

7. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex]

For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex]

R = 91.9 m for v 0 = 30 m/s; R = 163 m for v 0 ; R = 255 m for v 0 = 50 m/s

9. (a) 560 m/s (b) 800 × 10 3 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

11. 1.50 m, assuming launch angle of 45º

13. θ =6.1º. Yes, the ball lands at 5.3 m from the net

15. (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

17. 4.23 m. No, the owl is not lucky; he misses the nest.

19. No, the maximum range (neglecting air resistance) is about 92 m.

21. 15.0 m/s

23. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind.

25. [latex]y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{gt}^{2}=\left({v}_{0}\sin\theta\right)t-\frac{1}{2}{gt}^{2}\\[/latex] ,

so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]

[latex]x-{x}_{0}={v}_{0x}t=\left({v}_{0}\cos\theta \right)t=R\\[/latex], and substituting for t gives:

[latex]R={v}_{0}\cos\theta \left(\frac{{2v}_{0}\sin\theta}{g}\right)=\frac{{{2v}_{0}}^{2}\sin\theta \cos\theta }{g}\\[/latex]

since [latex]2\sin\theta \cos\theta =\sin 2\theta\\[/latex], the range is:

[latex]R=\frac{{{v}_{0}}^{2}\sin 2\theta }{g}\\[/latex].

Projectile Motion Problem Solving

Equations of motion for moving objects..

Notes/Highlights

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HONORS PHYSICS

Projectile Motion

PROJECTILE MOTION

Horizontal Projectiles

Angled projectiles.

General Strategy

Projectile motion problems, or problems of an object launched in both the x- and y- directions, can be analyzed using the physics you already know. The key to solving these types of problems is realizing that the horizontal component of the object’s motion is independent of the vertical component of the object’s motion. Since you already know how to solve horizontal and vertical kinematics problems, all you have to do is put the two results together!

Start these problems by making separate motion tables for vertical and horizontal motion. Vertically, the setup is the same for projectile motion as it is for an object in free fall. Horizontally, gravity only pulls an object down, it never pulls or pushes an object horizontally, therefore the horizontal acceleration of any projectile is zero. If the acceleration horizontally is zero, velocity must be constant, therefore v 0 horizontally must equal v horizontally. Finally, to tie the problem together, realize that the time the projectile is in the air vertically must be equal to the time the projectile is in the air horizontally.

When an object is launched or thrown completely horizontally, such as a rock thrown horizontally off a cliff, the initial velocity of the object is its initial horizontal velocity. Because horizontal velocity doesn’t change, this velocity is also the object’s final horizontal velocity, as well as its average horizontal velocity. Further, the initial vertical velocity of the projectile is zero. This means that you could hurl an object 1000 m/s horizontally off a cliff, and simultaneously drop an object off the cliff from the same height, and they will both reach the ground at the same time (even though the hurled object has traveled a greater distance).

Question: Fred throws a baseball 42 m/s horizontally from a height of 2m. How far will the ball travel before it reaches the ground? Answer: To solve this problem, you must first find how long the ball will remain in the air. This is a vertical motion problem. v 0 =0 m/s v=? Δy=2m a=9.8 m/s 2 t=? Now that you know the ball is in the air for 0.639 seconds, you can find how far it travels horizontally before reaching the ground. This is a horizontal motion problem, in which the acceleration is 0 (nothing is causing the ball to accelerate horizontally.) Because the ball doesn’t accelerate, its initial velocity is also its final velocity, which is equal to its average velocity. v 0 =42 m/s v=42 m/s Δx=? a=0 t=0.639s You can therefore conclude that the baseball travels 26.8 meters horizontally before reaching the ground. Question: Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30 meters per second. The time it takes projectile B to reach the level surface is: Answer: 3 seconds. They both take the same time to reach the ground because they both travel the same distance vertically, and they both have the same vertical acceleration (9.8 m/s 2 down) and initial vertical velocity (zero).

For objects launched at an angle, you have to do a little more work to determine the initial velocity in both the horizontal and vertical directions. For example, if a football is kicked with an initial velocity of 40 m/s at an angle of 30° above the horizontal, you need to break the initial velocity vector up into x- and y-components in the same manner as covered in the components of vectors math review section.

Then, use the components for your initial velocities in your horizontal and vertical tables. Finally, don't forget that symmetry of motion also applies to the parabola of projectile motion. For objects launched and landing at the same height, the launch angle is equal to the landing angle. The launch velocity is equal to the landing velocity. And if you want an object to travel the maximum possible horizontal distance (or range), launch it at an angle of 45°.

Question: Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground? Answer: Our first step in solving this type of problem is to determine Herman's initial horizontal and vertical velocity. We do this by breaking up his initial velocity into vertical and horizontal components: Next, we'll analyze Herman's vertical motion to find out how long he is in the air. We'll analyze his motion on the way up, find the time, and double that to find his total time in the air: v 0 =13 m/s v=0 Δy=? a y =-9.8 m/s 2 t=? Now that we know Herman was in the air 2.65s, we can find how far he moved horizontally, using his initial horizontal velocity of 22.5 m/s. v 0 =22.5 m/s v=22.5 m/s Δx=? a x =0 t=2.65s Therefore, Herman must have traveled 59.6m horizontally before returning to the Earth. Question: A golf ball is hit at an angle of 45° above the horizontal. What is the acceleration of the golf ball at the highest point in its trajectory? [Neglect friction.] Answer: 9.8 m/s 2 downward

IMAGES

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COMMENTS

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