Answer
$55.6^{\circ}$
Work Step by Step
It is given that:
initial speed of the ball $v_o=12\ m/s$
height $H=5\ m$
Maximum height of the projectile $H=\frac{v_o^2 sin^2\theta}{2g}$. Rearranging and substituting:
$sin^2\theta =\frac{2gh}{v_o^2}$
$sin\theta =\sqrt \frac{2gh}{v_o^2}$
$sin\theta =\sqrt \frac{2(9.8\ m/s^2)(5\ m)}{(12\ m/s)^2}$
$sin\theta =0.825$
$\theta =sin^{-1}(0.825)$
$\theta = 55.6^{\circ}$
