Answer
12.1 m
Work Step by Step
Let's consider the situation when the ball is thrown upward at an angle of $52^{\circ}$.
Here we use equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find its launch speed.
$\uparrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$0=(Vsin52)^{2}+2(-9.8\space m/s^{2})7.5\space m$
$V=\frac{12.12}{sin52^{\circ}}\space m/s=15.4\space m/s= Launch\space speed$
Let's consider the situation when the ball is thrown straight upward.
Here we use equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find its maximum height.
$\uparrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$0=(15.4\space m/s)^{2}+2(-9.8\space m/s^{2})S$
$S=12.1\space m$
Maximum height of the ball = 12.1 m
