## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 13

#### Answer

(a) $t = 0.064~s$ (b) The bullet's speed as it left the barrel was 780 m/s

#### Work Step by Step

(a) We can find the flight time of the bullet: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(0.020~m)}{9.80~m/s^2}}$ $t = 0.064~s$ (b) We can find the bullet's speed as it left the barrel; $v_x = \frac{x}{t} = \frac{50~m}{0.064~s}$ $v_x = 780~m/s$ The bullet's speed as it left the barrel was 780 m/s.

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