Projectile Motion Key Term projectile motion Two-Dimensional Motion Previously, we showed how quantities such as displacement and velocity were vectors that could be resolved into components. In this section, these components will be used to understand and predict the motion of objects thrown into the air. SECTION 3 Objectives Recognize examples of projectile motion. Describe the path of a projectile as a parabola. Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion. Use of components avoids vector multiplication. How can you know the displacement, velocity, and acceleration of a ball at any point in time during its flight? All of the kinematic equations could be rewritten in terms of vector quantities. However, when an object is propelled into the air in a direction other than straight up or down, the velocity, acceleration, and displacement of the object do not all point in the same direction. This makes the vector forms of the equations difficult to solve. One way to deal with these situations is to avoid using the complicated vector forms of the equations altogether. Instead, apply the technique of resolving vectors into components. Then you can apply the simpler one-dimensional forms of the equations for each component. Finally, you can recombine the components to determine the resultant. FIGURE 3.1 Motion of a Long Jumper When the long jumper is in the air, his velocity has both a horizontal and a vertical component. Components simplify projectile motion. When a long jumper approaches his jump, he runs along a straight line, which can be called the x-axis. When he jumps, as shown in Figure 3.1, his velocity has both horizontal and vertical components. Movement in this plane can be depicted by using both the x- and y-axes. Note that in Figure 3.2(b), a jumper s velocity vector is resolved into its two vector components. This way, the jumper s motion can be analyzed using the kinematic equations applied to one direction at a time. FIGURE 3.2 Michael Wong/Corbis Components of a Long Jumper s Velocity (a) A long jumper s velocity while sprinting along the runway can be represented by a horizontal vector. (b) Once the jumper is airborne, the jumper s velocity at any instant can be described by the components of the velocity. v (a) (b)
FIGURE 3.3 Air Resistance Affects Projectile Motion (a) Without air resistance, the soccer ball would travel along a parabola. (b) With air resistance, the soccer ball would travel along a shorter path. Path without air resistance Path with air resistance (b) (a) projectile motion the curved path that an object follows when thrown, launched, or otherwise projected near the surface of Earth In this section, we will focus on the form of two-dimensional motion called projectile motion. Objects that are thrown or launched into the air and are subject to gravity are called projectiles. Some examples of projectiles are softballs, footballs, and arrows when they are projected through the air. Even a long jumper can be considered a projectile. Projectiles follow parabolic trajectories. The path of a projectile is a curve called a parabola, as shown in Figure 3.3(a). Many people mistakenly think that projectiles eventually fall straight down in much the same way that a cartoon character does after running off a cliff. But if an object has an initial horizontal velocity, there will be horizontal motion throughout the flight of the projectile. Note that for the purposes of samples and exercises in this book, the horizontal velocity of projectiles will be considered constant. This velocity would not be constant if we accounted for air resistance. With air resistance, projectiles slow down as they collide with air particles, as shown in Figure 3.3(b). Did YOU Know? The greatest distance a regulation-size baseball has ever been thrown is 135.9 m, by Glen Gorbous in 1957. Projectile motion is free fall with an initial horizontal velocity. To understand the motion a projectile undergoes, first examine Figure 3.4 on the following page. The red ball was dropped at the same instant the yellow ball was launched horizontally. If air resistance is disregarded, both balls hit the ground at the same time. By examining each ball s position in relation to the horizontal lines and to one another, we see that the two balls fall at the same rate. This may seem impossible because one is given an initial velocity and the other begins from rest. But if the motion is analyzed one component at a time, it makes sense. First, consider the red ball that falls straight down. It has no motion in the horizontal direction. In the vertical direction, it starts from rest (v y,i = 0 m/s) and proceeds in free fall. Thus, the kinematic equations from the chapter Motion in One Dimension can be applied to analyze the vertical motion of the falling ball, as shown on the next page. Note that on Earth s surface the accel eration (a y ) will equal g ( 9.81 m/s 2 ) because the only vertical component of acceleration is free-fall acceleration. Note also that y is negative.
Vertical Motion of a Projectile That Falls from Rest v y,f = a y v y,f 2 = 2a y y y = 1 _ 2 a y ()2 Now consider the components of motion of the yellow ball that is launched in Figure 3.4. This ball undergoes the same horizontal displacement during each time interval. This means that the ball s horizontal velocity remains constant (if air resistance is assumed to be negligible). Thus, when the kinematic equations are used to analyze the horizontal motion of a projectile, the initial horizontal velocity is equal to the horizontal velocity throughout the projectile s flight. A projectile s horizontal motion is described by the following equation. FIGURE 3.4 Vertical Motion of a Projectile This is a strobe photograph of two table-tennis balls released at the same time. Even though the yellow ball is given an initial horizontal velocity and the red ball is simply dropped, both balls fall at the same rate. Horizontal Motion of a Projectile =,i = constant x = Next consider the initial motion of the launched yellow ball in Figure 3.4. Despite having an initial horizontal velocity, the launched ball has no initial velocity in the vertical direction. Just like the red ball that falls straight down, the launched yellow ball is in free fall. The vertical motion of the launched yellow ball is described by the same free-fall equations. In any time interval, the launched ball undergoes the same vertical displacement as the ball that falls straight down. For this reason, both balls reach the ground at the same time. To find the velocity of a projectile at any point during its flight, find the vector that has the known components. Specifically, use the Pythago rean theorem to find the magnitude of the velocity, and use the tangent function to find the direction of the velocity. PROJECTILE MOTION (bl) Richard Megna/Fundamental Photographs, New York; Roll a ball off a table. At the instant the rolling ball leaves the table, drop a second ball from the same height above the floor. Do the two balls hit the floor at the same time? Try varying the speed at which you roll the first ball off the table. Does varying the speed affect whether the two balls strike the ground at the same time? Next roll one of the balls down a slope. Drop the other ball from the base of the slope at the instant the first ball leaves the slope. Which of the balls hits the ground first in this situation? MATERIALS 2 identical balls slope or ramp SAFETY Perform this experiment away from walls and furniture that can be damaged.
Continued Projectiles Launched Horizontally Sample Problem D The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock s horizontal displacement is 45.0 m. Find the speed at which the rock was kicked. PREMIUM CONTENT Interactive Demo HMDScience.com ANALYZE Given: y = 321 m x = 45.0 m a y = g = 9.81 m/s 2 Unknown: v i = =? Diagram: The initial velocity vector of the rock has only a horizontal component. Choose the coordinate system oriented so that the positive y direction points upward and the positive x direction points to the right. 45.0 m 321 m PLAN Tips and Tricks The value for can be either positive or negative because of the square root. Because the object is moving in what has been selected as the positive direction, you choose the positive answer. SOLVE Choose an equation or situation: Because air resistance can be neglected, the rock s horizontal velocity remains constant. x = Because there is no initial vertical velocity, the following equation applies. y = 1_ 2 a y ()2 Rearrange the equations to isolate the unknowns: Note that the time interval is the same for the vertical and horizontal displacements, so the second equation can be rearranged to solve for. = _ 2 y a y Next rearrange the first equation for, and substitute the above value of into the new equation. = _ x a y ( = _ 2 y ) x Substitute the values into the equation and solve: = 9.81 m/s 2 (45.0 m) = 5.56 m/s (2) ( 321 m) CHECK YOUR ANSWER To check your work, estimate the value of the time interval for x and solve for y. If is about 5.5 m/s and x = 45 m, 8 s. If you use an approximate value of 10 m/s 2 for g, y 320 m, almost identical to the given value.
Projectiles Launched Horizontally (continued) 1. A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling? 2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed? 3. A pelican flying along a horizontal path drops a fish from a height of 4 m. The fish travels 8.0 m horizontally before it hits the water below. What is the pelican s speed? 4. If the pelican in item 3 was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below? Use components to analyze objects launched at an angle. Let us examine a case in which a projectile is launched at an angle to the horizontal, as shown in Figure 3.5. The projectile has an initial vertical component of velocity as well as a horizontal component of velocity. Suppose the initial velocity vector makes an angle θ with the horizontal. Again, to analyze the motion of such a projectile, you must resolve the initial velocity vector into its components. The sine and cosine functions can be used to find the horizontal and vertical components of the initial velocity.,i = v i cos θ and v y,i = v i sin θ We can substitute these values for,i and v y,i into the kinematic equations to obtain a set of equations that can be used to analyze the motion of a projectile launched at an angle. FIGURE 3.5 Components of Initial Velocity An object is projected with an initial velocity, v i, at an angle of θ. Resolve the initial velocity into its x and y components. Then, the kinematic equations can be applied to describe the motion of the projectile throughout its flight. v i Projectiles Launched at an Angle =,i = v i cos θ = constant x = (v i cos θ) v y,f = v i sin θ + a y v 2 y,f = v 2 i (sin θ) 2 + 2a y y y = (v i sin θ) + _ 1 2 a y ()2 As we have seen, the velocity of a projectile launched at an angle to the ground has both horizontal and vertical components. The vertical motion is similar to that of an object that is thrown straight up with an initial velocity.
Projectiles Launched at an Angle Sample Problem E A zookeeper finds an escaped monkey on a pole. While aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the pole, which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the moment the zookeeper shoots, the monkey drops a banana. The dart travels at 50.0 m/s. Will the dart hit the monkey, the banana, or neither one? ANALYZE Select a coordinate system. The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m. 10.0 m 4.00 m 1.00 m PLAN SOLVE Use the inverse tangent function to find the angle of the dart with the x-axis. θ = tan ( 1 _ y x ) = ( tan 1 _ 4.00 m 10.0 m ) = 21.8 Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate, the unknown, the time the dart takes to travel the horizontal distance. x = (v i cos θ) = _ x v i cos θ = 10.0 m (50.0 m/s)(cos 21.8 ) = 0.215 s Find out how far each object will fall during this time. Use the free-fall kinematic equation. For the banana, v i = 0. Thus: y b = _ 1 2 a y ()2 = _ 1 2 ( 9.81 m/s2 )(0.215 s) 2 = 0.227 m The dart has an initial vertical component of velocity of v i sin θ, so: y d = (v i sin θ) + _ 1 2 a y ()2 y d = (50.0 m/s)(sin 21.8 )(0.215 s) + _ 1 2 ( 9.81 m/s2 )(0.215 s) 2 y d = 3.99 m 0.227 m = 3.76 m Find the final height of both the banana and the dart. y b, f = y b,i + y b = 5.00 m + ( 0.227 m) = 4.77 m above the ground y d, f = y d,i + y d = 1.00 m + 3.76 m = 4.76 m above the ground The dart hits the banana. The slight difference is due to rounding.
Projectiles Launched at an Angle (continued) 1. In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0 m away. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15 with respect to the flat roof. Will he make it to the other roof, which is 2.5 m lower than the building he jumps from? 2. A golfer hits a golf ball at an angle of 25.0 to the ground. If the golf ball covers a horizontal distance of 301.5 m, what is the ball s maximum height? (Hint: At the top of its flight, the ball s vertical velocity component will be zero.) 3. A baseball is thrown at an angle of 25 relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high above the thrower did the ball travel? 4. Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00 m from a waterfall that is 0.55 m tall and jumps at an angle of 32.0. What must be the salmon s minimum speed to reach the waterfall? SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Which of the following exhibit parabolic motion? a. a flat rock skipping across the surface of a lake b. a three-point shot in basketball c. a space shuttle while orbiting Earth d. a ball bouncing across a room e. a life preserver dropped from a stationary helicopter 2. During a thunderstorm, a tornado lifts a car to a height of 125 m above the ground. Increasing in strength, the tornado flings the car horizontally with a speed of 90.0 m/s. How long does the car take to reach the ground? How far horizontally does the car travel before hitting the ground? Interpreting Graphics 3. An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as illustrated in Figure 3.6. The plane is traveling horizontally at 30.0 m/s at a height of 200.0 m above the ground. a. What horizontal distance does the package fall before landing? b. Find the velocity of the package just before it hits the ground. FIGURE 3.6 Dropping a Package plane = 30.0 m/s 200.0 m