## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 150: 44

#### Answer

(a) The ball will rise 18.5 meters above the initial height. (b) The total time of flight is $3.89~s$ (c) The fielder is a distance of 42.8 meters from home plate.

#### Work Step by Step

(a) We can find the maximum vertical displacement above the initial height: $v_{fy}^2 = v_{0y}^2+2a\Delta y$ $\Delta y = \frac{v_{fy}^2 - v_{0y}^2}{2a}$ $\Delta y = \frac{0 -[(22.0~m/s)~sin~60.0^{\circ}]^2}{(2)(-9.80~m/s^2)}$ $\Delta y = 18.5~m$ The ball will rise 18.5 meters above the initial height. (b) We can find the time it takes to reach the maximum height: $v_{fy} = v_{0y}+a_y~t$ $t = \frac{v_{fy}-v_{0y}}{a_y}$ $t = \frac{0-(22.0~m/s)~sin~60.0^{\circ}}{-9.80~m/s^2}$ $t = 1.944~s$ The total time of flight is twice this time. The total time of flight is $(2)(1.944~s)$ which is $3.89~s$ (c) We can find the horizontal displacement: $\Delta x = v_x~t = (22.0~m/s)~cos~60.0^{\circ}~(3.89~s) = 42.8~m$ The fielder is a distance of 42.8 meters from home plate.

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