## 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: 44

#### Answer

(a) The cliff is 26 meters high. (b) The maximum height is 34 meters. (c) The impact speed is 20 m/s

#### Work Step by Step

(a) $y = v_{0y}t-\frac{1}{2}gt^2$ $y = (30~m/s)~sin(60^{\circ})(4.0~s)-\frac{1}{2}(9.80~m/s^2)(4.0~s)^2$ $y = 26~m$ The cliff is 26 meters high. (b) $2ah=v_y^2-v_{0y}^2$ $2(-g)h=v_y^2-v_{0y}^2$ $h=\frac{v_{0y}^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2}{2g}$ $h=\frac{[(30~m/s)~sin(60^{\circ})]^2}{(2)(9.80~m/s^2)}$ $h = 34~m$ The maximum height is 34 meters. (c) We can find the horizontal component of velocity. $v_x = v_0~cos(\theta)$ $v_x = (30~m/s)~cos(60^{\circ})$ $v_x = 15~m/s$ We can find the vertical component of velocity after 4.0 seconds. $v_y = v_{0y}+at$ $v_y = (30~m/s)~sin(60^{\circ})-(9.80~m/s^2)(4.0~s)$ $v_y = -13.2~m/s$ We can find the impact speed. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(15~m/s)^2+(-13.2~m/s)^2}$ $v = 20~m/s$ The impact speed is 20 m/s.

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