HW04 - Projectile Motion

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 1/24 HW04 - Projectile Motion Due: 11:59pm on Friday, October 14, 2022 To understand how points are awarded, read the Grading Policy for this assignment. Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components , you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often-used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity → v v_vec . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more manageable, it is common to break up such a quantity into its x component v x (v_x) and its y component v y (v_y) . Consider a particle with initial velocity → v v_vec that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis. Part A What is the x component v x v_x of → v v_vec ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component v y v_y of → v v_vec ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C ( ) ( ) v x v_x = -6.00 m/s v y v_y = 10.4 m/s

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 2/24 Click on the image below to launch the video: Projectile Motion . Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x -component motion diagram is what you would get if you shone a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shone a spotlight to the left and recorded the particle's shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude g = 10 m/s 2 . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation y = y 0 + v 0 t + (1/2) at 2 . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation x = x 0 + v 0 t . Now, consider a variation on this problem in which two balls are simultaneously dropped from a height of 5.0 m. Click on the image below to launch the second video: Projectile Motion . Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity.

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 4/24 Part A Find the time t H t_H it takes the projectile to reach its maximum height. Express t H t_H in terms of v 0 v_0 , θ theta , and g g (the magnitude of the acceleration due to gravity). Hint 1. A basic property of projectile motion It is important to understand that the motion in the y direction is independent of the motion in the x direction because these directions are orthogonal (at right angles) to each other. Hint 2. What condition applies at the top? What is the value of the projectile's vertical velocity when it has reached the top of its trajectory? ANSWER: Hint 3. Vertical velocity as a function of time Find an expression for the projectile's vertical velocity as a function of time, v y ( t ) . Express your answer in terms of t t , v 0 v_0 , θ theta , and g g , the acceleration due to gravity. Hint 1. Initial velocity in y direction What is the initial y component of velocity? Provide your answer in terms of v 0 v_0 and θ theta . ANSWER: ANSWER: v top v_top = 0 v y0 v_y0 = v 0 sin( θ ) v y ( t ) = v 0 sin( θ ) − gt

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 5/24 Hint 4. Putting it all together You now have a general expression for the y-velocity as a function of time, and you also know the the y-velocity at the top (i.e. at time t H t_H ). This gives an equation which you can solve for t H t_H . Hint 5. A list of possible answers There are four answers in the pulldown list below, one of which is the correct answer for this question (the time to reach the top). Once you have finished this part, therefore, you will have the answer for the question. (However, you will still need to enter this answer into the answer box in the main part of the problem.) ANSWER: ANSWER: Correct Part B Find t R t_R , the time at which the projectile hits the ground. Express the time in terms of v 0 v_0 , θ theta , and g g . Hint 1. Two possible approaches There are two good ways to find the total flight time t R t_R : either by invoking the symmetry of this problem (limited to projectiles fired over level ground) or by finding a general expression for y ( t )y(t) , the y position of the projectile, and setting this equal to the height at the end of the trajectory, y R y_R . The second method is more general (i.e., it would work if the projectile landed on a hill of height H). What is the value of y R y_R in this problem? Express y R y_R in terms of quantities given in the introduction. ANSWER: Hint 2. Some needed kinematics Give an expression for the height as a function of time, y ( t )y(t) , taking y 0 = 0 . v 0 g sin ( θ ) v 0 g sin ( θ ) v 0 sin ( θ ) g g sin ( θ ) v 0 t H t_H = v 0 sin ( θ ) g y R y_R = 0

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 7/24 Hint 3. Finding H H Remember that H = y ( t max ) . ANSWER: Correct Part D Find the total distance R R (often called the range) traveled in the x direction; in other words, find where the projectile lands. Express the range in terms of v 0 v_0 , θ theta , and g g . Hint 1. When does the projectile hit the ground? The projectile reaches the ground at time t R t_R . Hint 2. Where is the projectile as a function of time? Give an expression for the x position of the particle as a function of time. Answer in terms of t t , v 0 v_0 , and θ theta . Hint 1. Acceleration in x direction There is no acceleration in the x direction. ANSWER: Hint 3. Finding the range Remember that R = x ( t R ) . Hint 4. A list of possible answers Choose the correct answer for R R from the following list. (You will still have to enter this answer into the main answer box to receive credit.) ANSWER: t max t_max = v 0 sin ( θ ) g H ( θ )H(theta) = 1 2 g v 0 sin( θ ) 2 ( ) x ( t )x(t) = v 0 cos( θ ) t

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 8/24 ANSWER: Correct The actual formula for R ( θ )R(theta) is less important than how it is obtained: 1. Consider the x and y motion separately. 2. Find the time of flight from the y -motion 3. Find the x -position at the end of the flight - this is the range. If you remember these steps, you can deal with many variants of the basic problem, such as: a cannon on a hill that fires horizontally (i.e. the second half of the trajectory), a projectile that lands on a hill, or a projectile that must hit a moving target. Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: 1. An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, v x v_x , is constant. 2. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by g g , is equal to 9.80 m/s 2 near the surface of the earth. Hence, the y component of its velocity, v y v_y , changes continuously. 3. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time t t the projectile is in the air. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t 0 = 0 s corresponds to the moment just after the ball is launched from position x 0 = 0 m and y 0 = 0 m . Its launch velocity, also called v 2 0 g sin ( 2 θ ) v 0 sin ( θ ) g v 2 0 g cos ( 2 θ ) v 2 0 sin ( 2 θ ) g R ( θ )R(theta) = v 0 2 g sin(2 θ )

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 10/24 What are the values of the intial velocity vector components v 0 , x and v 0 , y (both in m/s ) as well as the acceleration vector components a 0 , x and a 0 , y (both in m/s 2 )? Here the subscript 0 means "at time t 0 t_0 ." Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector → v 0 v_0_vec points up and to the right. Since "up" is the positive y axis direction and "to the right" is the positive x axis direction, v 0 , x and v 0 , y will both be positive. As shown in the figure, v 0 , x , v 0 , y , and v 0 v_0 are three sides of a right triangle, one angle of which is θ theta . Thus v 0 , x and v 0 , y can be found using the definition of the sine and cosine functions given below. Recall that v 0 = 30.0 m/s and θ = 60.0 degrees and note that sin( θ ) = length of opposite side length of hypotenuse = v 0 , y v 0 , cos( θ ) = length of adjacent side length of hypotenuse = v 0 , x v 0 . What are the values of v 0 , x and v 0 , y ? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: Correct Also notice that at time t 2 t_2 , just before the ball lands, its velocity components are v 2 , x = 15 m/s (the same as always) and v 2 , y = − 26.0 m/s (the same size but opposite sign from v 0 , y by symmetry). The acceleration at time t 2 t_2 will have components (0, -9.80 m/s 2 ), exactly the same as at t 0 t_0 , as required by Rule 2. The peak of the trajectory occurs at time t 1 t_1 . This is the point where the ball reaches its maximum height y max y_max . At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. 15.0,26.0 m/s 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 11/24 Part C What are the values of the velocity vector components v 1 , x and v 1 , y (both in m/s ) as well as the acceleration vector components a 1 , x and a 1 , y (both in m/s 2 )? Here the subscript 1 means that these are all at time t 1 t_1 . ANSWER: Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched ( t 0 t_0 ) until just before it lands ( t 2 t_2 ). Hence the flight time can be calculated as t 2 − t 0 , or just t 2 t_2 in this particular situation since t 0 = 0 . Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height y max y_max , how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball off a cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity ( v 0 , y = 0 ) as well as the same acceleration ( → a = g downward). They differ only in their x velocity (one is zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 13/24 Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration g g is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. Problem 4.15 - Enhanced - with Expanded Hints A rifle is aimed horizontally at a target 48.0 m away. The bullet hits the target 2.10 cm below the aim point. Part A What was the bullet's flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Treat the bullet as a particle and neglect the influence of air resistance. The bullet's motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Use the appropriate kinematic equation to find the flight time of the bullet. Hint 2. Simplify: kinematic equations Which of the following equations is most applicable for determining the flight time of the bullet? ANSWER: Increase v 0 v_0 above 30 m/s . Reduce v 0 v_0 below 30 m/s . Reduce θ theta from 60 degrees to 45 degrees . Reduce θ theta from 60 degrees to less than 30 degrees . Increase θ theta from 60 degrees up toward 90 degrees . v f x = v i x + a x Δ t y f = y i + v i y Δ t + 1 2 a y (Δ t ) 2 v f y = v i y + a y Δ t x f = x i + v i x Δ t + 1 2 a x (Δ t ) 2

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 14/24 ANSWER: Correct MODEL: The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is neglected. VISUALIZE: Known x 0 = y 0 = t 0 = 0 v 0 y = 0 x 1 = 48.0 m y 1 = − 2.10 cm a y = − g Find t 1 , v 0 x SOLVE: Using y 1 = y 0 + v 0 y ( t 1 − t 0 ) + 1 2 a y ( t 1 − t 0 ) 2 , we obtain ( − 2.10 × 10 − 2 m) = 0 m + 0 m + 1 2 ( − 9.8 m/s 2 )( t 1 − 0 s) 2 t 1 = 0.0655 s Part B What was the bullet's speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem Treat the bullet as a particle and neglect the influence of air resistance. The bullet's motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. After determining the flight time of the bullet, use the appropriate kinematic equation to find its initial speed. Hint 2. Simplify: kinematic equations Which of the following equations is most applicable for determining the bullet's speed as it left the barrel? ANSWER: t = 6.55×10 −2 s

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 16/24 Hint 2. Find v y 0 What is the vertical component v y0 v_y0 of the initial velocity? Express your answer symbolically in terms of v x0 v_x0 . ANSWER: ANSWER: Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find t a t_a in terms of v y0 v_y0 and g g . Give your answer in terms of v y0 v_y0 and g g . Hint 1. Find v y t a v_y(t_a) When applied to the y-component of velocity, in this problem the formula for v ( t )v(t) with constant acceleration − g -g is v y ( t ) = v y 0 − g t What is v y t a v_y(t_a) , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of v y0 v_y0 only. ANSWER: ANSWER: Hint 3. Put the algebra together to find t a t_a symbolically. If you have an expression for the initial vertical velocity component in terms in terms of D D and t a t_a , and another in terms of g g and t a t_a , you should be able to eliminate this initial component to find an expression for t 2 a t^2_a Express your answer symbolically in terms of given variables. v x 0 = D t a v y 0 = v x0 v y 0 = D t a ( ) ( ) v y ( t a ) = − v y0 t a = 2 v y0 g

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 17/24 ANSWER: ANSWER: Correct Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: ANSWER: Correct Problem 4.57 - Enhanced - with Hints and Feedback A stunt man drives a 1500 kg car at a speed of 20 m/s off a 20- m -high cliff. The road leading to the cliff is inclined upward at an angle of 20 . t 2 a = 2 D g t a = 6.7 s t f = 1.1 s t d = 5.6 s

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 19/24 Hint 1. How to approach the problem Start by drawing a picture of the golf ball's path, showing its starting and ending points. Choose a coordinate system, and label the origin (it is conventional to let x be the horizontal direction and y the vertical direction). Some information you should note are the x and y components of the initial velocity as well as the acceleration. Next, apply the equations for x x and y y as a function of time, x ( t )x(t) and y ( t )y(t) . Given the final position of the ball upon hitting the ground, you should be able to combine these two equations to solve for the total distance traveled in each case. Hint 2. Simplify: distance equation Using the kinematic equation for the vertical position of the ball, you can find the expression for its flight time and then insert it into the equation for the ball's horizontal position. Derive the expression for the horizontal distance traveled in terms of the ball's initial velocity v 0 v_0 , the launch angle θ theta , and the free-fall acceleration g g . Express your answer in terms of v 0 v_0 , θ theta , and g g . ANSWER: Hint 3. Simplify: distance traveled on the moon Calculate the horizontal distance traveled by the ball on the moon. Express your answer with the appropriate units. ANSWER: ANSWER: Δ x = 2 v 0 2 sin ( θ ) cos ( θ ) g (Δ x ) moon = 53.0 m L = 44 m

10/20/22, 12:29 AM HW04 - Projectile Motion https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072421 20/24 Correct MODEL: The golf ball is a particle following projectile motion. VISUALIZE: Known x 0 = y 0 = t 0 = 0 v 0 = 10 m/s θ theta = 30 a x = 0 a y = − g g earth = 9.8 m/s 2 g moon = g earth /6 Find x 1 , t 1 SOLVE: The distance traveled is x 1 = v 0 x t 1 = v 0 cos θ t 1 . The flight time is found from the y -equation, using the fact that the ball starts and ends at y = 0: y 1 − y 0 = 0 = v 0 sin θ t 1 − 1 2 gt 2 1 = ( v 0 sin θ − 1 2 gt 1 ) t 1 t 1 = 2 v 0 sin θ g Thus the distance traveled is x 1 = v 0 cos θ × 2 v 0 sin θ g = 2 v 2 0 sin θ cos θ g For θ theta = 30 , the distances are ( x 1 ) earth = 2 v 2 0 sin θ cos θ g earth = 2 ( 10 m / s ) 2 sin 30 cos 30 9.80 m / s 2 = 8.8 m ( x 1 ) moon = 2 v 2 0 sin θ cos θ g moon = 2 v 2 0 sin θ cos θ g earth / 6 = 6 × 2 v 2 0 sin θ cos θ g earth = 6( x 1 ) earth = 53.0 m The golf ball travels 53.0 m − 8.8 m = 44.2 m ≈ 44 m farther on the moon than on earth. Part B For how much more time was the ball in flight? Express your answer with the appropriate units. Hint 1. How to approach the problem Now that you know the horizontal range, you can use either of your equations from the previous part, x ( t ) and y ( t ) , to solve for the time. Hint 2. Simplify: time equation Given that the initial and final positions of the ball are at the same height, derive the expression for the flight time in terms of the ball's initial velocity v 0 v_0 , the launch angle θ theta , and the free-fall acceleration g g . Express your answer in terms of v 0 v_0 , θ theta , and g g . ANSWER: