College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 149: 40

Answer

(a) After 1.60 seconds have elapsed, the ball is 1.46 meters above the ground a horizontal distance of 32.0 meters away. (b) The ball will hit the ground after another 0.09 seconds have elapsed. At this time, the ball will be a horizontal distance of 33.8 meters away.

Work Step by Step

(a) We can find the vertical position of the ball after $1.60~s$: $y = y_0+\frac{1}{2}at^2$ $y = 14.0~m+\frac{1}{2}(-9.80~m/s^2)(1.60~s)^2$ $y = 1.46~m$ We can find the horizontal position after $1.60~s$: $x = v_x~t$ $x = (20.0~m/s)(1.60~s)$ $x = 32.0~m$ After 1.60 seconds have elapsed, the ball is 1.46 meters above the ground a horizontal distance of 32.0 meters away. (b) We can find the time it takes for the ball to hit the ground from a height of $14.0~m$: $\Delta y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2\Delta y}{a}}$ $t = \sqrt{\frac{(2)(-14.0~m)}{-9.80~m/s^2}}$ $t = 1.69~s$ We can find the horizontal position after $1.69~s$: $x = v_x~t$ $x = (20.0~m/s)(1.69~s)$ $x = 33.8~m$ The ball will hit the ground after another 0.09 seconds have elapsed. At this time, the ball will be a horizontal distance of 33.8 meters away from the starting point.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.