Answer
62.85 m/s
Work Step by Step
Let's apply equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find the initial velocity (V) of the projectile.
$\uparrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$0=(Vsin15^{\circ})^{2}+2(-9.8\space m/s^{2})13.5\space m$
Launch speed = 62.85 m/s
$V=\frac{\sqrt {264.6}}{sin15^{\circ}}m/s=62.85\space m/s$
