Projectile Motion

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.

Learning Objectives By the end of this section, you will be able to:. Use one-dimensional motion in perpendicular directions to analyze projectile motion. Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface. Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch. Calculate the trajectory of a projectile.Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity.

The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory.

The motion of falling objects as discussed in 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, and our treatment neglects the effects of air resistance.The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately.

We discussed this fact in, where we saw that vertical and horizontal motions are 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 resulting from gravity is vertical; thus, there is 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. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes.

Illustrates the notation for displacement, where we define s → s → to be the total displacement, and x → x → and y → y → are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

Figure 4.11 The total displacement s of a soccer ball at a point along its path. The vector s → s → has components x → x → and y → y → along the horizontal and vertical axes.

Its magnitude is s and it makes an angle Φ with the horizontal.To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple. A y = − g = −9.8 m / s 2 ( − 32 ft / s 2 ). A y = − g = −9.8 m / s 2 ( − 32 ft / s 2 ).Because gravity is vertical, a x = 0.

If a x = 0, a x = 0, this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0 x. With these conditions on acceleration and velocity, we can write the kinematic through for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from. The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = − g, a x = 0: a y = − g, a x = 0:Horizontal Motion. Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the components of displacement s → s → along these axes are x and y. The magnitudes of the components of velocity v → v → are v x = v cos θ and v y = v sin θ, v x = v cos θ and v y = v sin θ, where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in. Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical.

Use the kinematic equations for horizontal and vertical motion presented earlier. 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 those for one-dimensional kinematics and are illustrated in the following solved examples. Recombine quantities in the horizontal and vertical directions to find the total displacement s → s → and velocity v →. Solve for the magnitude and direction of the displacement and velocity using. Figure 4.12 (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 a x = 0 a x = 0 and v x v x is a 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 toward 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 4.7A Fireworks Projectile Explodes High and AwayDuring a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0 ° 75.0 ° above the horizontal, as illustrated in. 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 passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point? Figure 4.13 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.StrategyThe motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and a y = − g. We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the desired quantities.Solution(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. Y = 233 m.Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from 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 reaches 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, so the initial velocity would have to be somewhat larger than that given to reach the same height.(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0 y − g t.

V y = v 0 y − g t. Because v y = 0 v y = 0 at the apex, this equation reduces to simply. T = v 0 y g = 67.6 m / s 9.80 m / s 2 = 6.90 s. T = v 0 y g = 67.6 m / s 9.80 m / s 2 = 6.90 s.This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + 1 2 ( v 0 y + v y ) t.

Y = y 0 + 1 2 ( v 0 y + v y ) t. This is left for you as an exercise to complete.(c) Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the 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. X = ( 18.1 m / s ) 6.90 s = 125 m. X = ( 18.1 m / s ) 6.90 s = 125 m.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. When the shell explodes, air resistance has a major effect, and many fragments land directly below.(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point. Figure 4.14 The trajectory of a tennis ball hit into the stands.StrategyAgain, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity.

This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain v → v → at final time t, determined in the first part of the example.Solution(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using. T = 3.79 s.The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.(b) We can find the final horizontal and vertical velocities v x v x and v y v y with the use of the result from (a).

Then, we can combine them to find the magnitude of the total velocity vector v → v → and the angle θ θ it makes with the horizontal. Since v x v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Θ v = tan −1 ( v y v x ) = tan −1 ( −15.9 21.2 ) = 36.9 °. Θ v = tan −1 ( v y v x ) = tan −1 ( −15.9 21.2 ) = 36.9 °.Significance(a) As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.

(b) The negative angle means the velocity is 36.9 ° 36.9 ° below the horizontal at the point of impact. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation. Time of Flight, Trajectory, and RangeOf interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flightWe can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find. (4.24)This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Does not apply when the projectile lands at a different elevation than it was launched, as we saw in of the tennis player hitting the ball into the stands.

The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Stand o food game play online. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long. TrajectoryThe trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y( x).

We take x 0 = y 0 = 0 x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives. (4.26)Note particularly that is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 v 0 and sin 2 θ 0 sin 2 θ 0, and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity.

Furthermore, we see from the factor sin 2 θ 0 sin 2 θ 0 that the range is maximum at 45 °. These results are shown in. In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45 °.

This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90 °. The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Example 4.9Comparing Golf ShotsA golfer finds himself in two different situations on different holes. On the second hole he is 120 m from the green and wants to hit the ball 90 m and let it run onto the green.

He angles the shot low to the ground at 30 ° 30 ° to the horizontal to let the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at 70 ° 70 ° to the horizontal to minimize rolling after impact. Both shots are hit and impacted on a level surface.(a) What is the initial speed of the ball at the second hole?(b) What is the initial speed of the ball at the fourth hole?(c) Write the trajectory equation for both cases.(d) Graph the trajectories.StrategyWe see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a) and (b). Figure 4.16 Two trajectories of a golf ball with a range of 90 m. The impact points of both are at the same level as the launch point.SignificanceThe initial speed for the shot at 70 ° 70 ° is greater than the initial speed of the shot at 30 °.

Note from that two projectiles launched at the same speed but at different angles have the same range if the launch angles add to 90 °. The launch angles in this example add to give a number greater than 90 °. Thus, the shot at 70 ° 70 ° has to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance. If the two golf shots in were launched at the same speed, which shot would have the greatest range?When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in, which is based on a drawing in Newton’s Principia.

If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit.

These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in. Want to cite, share, or modify this book? This book is Creative Commons Attribution License4.0 and you must attribute OpenStax.Attribution information.If you are redistributing all or part of this book in a print format,then you must include on every physical page the following attribution:Access for free at you are redistributing all or part of this book in a digital format,then you must include on every digital page view the following attribution:Access for free atCitation information.Use the information below to generate a citation.

We recommend using acitation tool such as.Authors: William Moebs, Samuel J. Ling, Jeff Sanny.Publisher/website: OpenStax.Book title: University Physics Volume 1.Publication date: Sep 19, 2016.Location: Houston, Texas.Book URL:.Section URL:© Jan 30, 2020 OpenStax.

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