## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 61: 2.36

#### Answer

(a) The speed of the rock just before it hits the street is 32.7 m/s (b) The time that elapses is 5.6 seconds.

#### Work Step by Step

Let up be the positive direction. Let $y_0 = +30.0~m$ (a) Let $v$ be the speed of the rock just before it hits the street. $v^2 = v_0^2 + 2a(y-y_0)$ $v = \sqrt{v_0^2 + 2a(y-y_0)}$ $v = \sqrt{(22.0~m/s)^2 + (2)(-9.80~m/s^2)(-30.0~m)}$ $v = 32.7~m/s$ The speed of the rock just before it hits the street is 32.7 m/s (b) $t = \frac{v-v_0}{a} = \frac{-32.7~m/s-22.0~m/s}{-9.80~m/s^2}$ $t = 5.6~s$ The time that elapses is 5.6 seconds.

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