Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 53

Answer

The launch speed is 7.4 m/s

Work Step by Step

We can find the time it takes for the steel ball to fall 2.5 meters. $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(2.5~m)}{9.80~m/s^2}}$ $t = 0.714~s$ We can find the horizontal component of the plastic ball's speed. $v_x = \frac{x}{t} = \frac{4.0~m}{0.714~s}$ $v_x = 5.6~m/s$ We can find the vertical component of the plastic ball's launch speed. $y = v_{0y}t-\frac{1}{2}gt^2$ $v_{0y} = \frac{y+\frac{1}{2}gt^2}{t}$ $v_{0y} = \frac{1.0~m+\frac{1}{2}(9.80~m/s^2)(0.714~s)^2}{0.714~s}$ $v_{0y} = 4.9~m/s$ We can find the launch speed. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(5.6~m/s)^2+(4.9~m/s)^2}$ $v = 7.4~m/s$ The launch speed is 7.4 m/s
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.