Chapter 4: Two-Dimensional KinematicsAnswers to Even-Numbered Conceptual Questions2. The y component of velocity is first positive and then negative in a symmetric fashion. As a result, theaverage y component of velocity is zero. The x component of velocity, on the other hand, is always v0cos θ.Therefore, the projectile’s average velocity has a magnitude of v0cosθ and points in the positive x direction.4. (a) No. The acceleration is always vertically downward, but the fly ball is always moving at an angle to thevertical, never straight down. Therefore, its velocity is never vertical and is never parallel to the acceleration.(b) Yes. A projectile at the top of its trajectory has a velocity that is horizontal, while at the same time itsacceleration is vertical.6. Just before it lands, this projectile is moving downward with the same speed it had when it was launched. Inaddition, if it was launched upward at an angle θ above the x axis, it is moving in a direction that is an angleθ below the x axis just before it lands. Therefore, its velocity just before landing is rv = 2m/s( )ˆx+ – 4m/s( )ˆy.8. Maximum height depends on the initial speed squared. Therefore, to reach twice the height, projectile 1must have an initial speed that is the square root of 2 times greater than the initial speed of projectile 2. Itfollows that the ratio of the speeds is the square root of 2.10. The tomato lands on the road in front of you. This follows from the fact that its horizontal speed is the sameas yours during the entire time of its fall.Solutions to Problems and Conceptual Exercises1. Picture the Problem: You walk briskly down the street while tossing a ball in the air and catching it again.Strategy: Use a separate analysis of the horizontal and vertical motions of the ball to answer the conceptual question.Solution: 1. (a) As long as air friction is neglected there is no acceleration of the ball in the horizontal direction. The ball will continue moving horizontally with the same speed as your walking speed. Therefore, you need to launch the ball straight upward relative to yourself in order for it to land back in your hand.2. (b) The best explanation (see above) is III. The ball moves in the forward direction with your walking speed at all times. Statements I and II are false because they ignore the inertia of the ball in the horizontal direction.Insight: If air friction is taken into account you must launch the ball in the forward direction a little bit. While it is in the air the friction will slow the ball horizontally so that it lands back in your hand.2. Picture the Problem: The vector involved in this problem is depicted at right.Strategy: Separate rv into x- and y-components. Let north be along the y-axis, west along the −x axis. Find the components of the velocity in each direction and use them to find the distances traveled.Solution: 1. (a) Find vx and vy: vx=− 4.2 m/s( )cos32°=−3.56 m/svy= 4.2 m/s( )sin32°=2.23 m/s2. Find the westwarddistance traveled: y =vxt = 3.56 m/s( )25 min × 60 s/min( )=5300 m= 5.3 km3. (b) Find the northward distance traveled: x =vyt = 2.23 m/s( )25 min×60 s/min( )=3300 m= 3.3 kmCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.4 – 1NxvvrW32°yvChapter 4: Two-Dimensional Kinematics James S. Walker, Physics, 4th EditionInsight: The northward and westward motions can be considered separately. In this case they are each described by constant velocity motion.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.4 – 2Chapter 4: Two-Dimensional Kinematics James S. Walker, Physics, 4th Edition3. Picture the Problem: The vectors involved in this problem are depicted at right.Strategy: Let north be along the y-axis and east along the x-axis. Find the components of the velocity in each direction, and use them to find the times elapsed.Solution: 1. (a) Find the x component of rv: vx= 1.75 m/s( )cos18.0°=1.66 m/s2. Find the time elapsed to travel east 20.0 m: t =xvx=20.0 m1.66 m/s=12.0 s3. (b) Find the y component of rv: vy= 1.75 m/s( )sin18.0°=0.541 m/s4. Find the time elapsed to travel north 30.0 m: t =yvy=30.0 m0.541 m/s= 55.5 sInsight: The northward and eastward motions can be considered separately. In both cases the actual distance traveled isgreater than 20.0 or 30.0 m, respectively. For instance, in the second case you must actually travel a total distance of 30.0 m sin18.0° =97.1 m to change your displacement by 30.0 m north.4. Picture the Problem: The car moves up the 5.5° incline with constant acceleration, changing both its horizontal and vertical displacement simultaneously.Strategy: Find the magnitude of the displacement along the incline, and then independently find the horizontal and vertical components of the displacement.Solution: 1. (a) Find the magnitude of the displacement along the incline using equation 2-11: Δr =0+v0t +12at2=0+0+122.0 m/s2( )12 s( )2=140 m2. Find the horizontal component of Δr: x =dcosθ = 140 m( )cos5.5°=140 m3. (b) Find the vertical component Δr: y =dsinθ = 140 m( )sin5.5°=14 mInsight: The horizontal and vertical motions can be considered separately. In this case they are each described by constant acceleration motion, but the vertical acceleration is less than the horizontal. The two accelerations would be equal if the angle of the incline were 45°.5. Picture the Problem: The motion of the particle is depicted at right.Strategy: Use the given information to independently write the equations of motion in the x and y directions. There will be a pair of equations for the position of the particle (like equation 4-6) and a pair for the velocity (like equation 4-6), except in this case the acceleration will not be the same as rg.Solution: 1. (a) Write equation 4-6 for the x position of the particle using ra = −4.4 m/s2( )ˆx instead of rg: x =x0+v0xt +12axt2=0+0+12−4.4 m/s2( )(5.0)2x = –55 m2. Now do the same for the y coordinate of the position: y =y0+v0yt +12ayt2=0+ 6.2 m/s( )(5.0 s)+0= 31 m3. (b) Write equation 4-6 for