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.
