Answer
17.9 m/s
Work Step by Step
Here we use equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find the launch speed (V) of the skier.
$\uparrow V^{2}=u^{2}+2aS$
$0^{2}=(Vsin63^{\circ})^{2}+2\times(-9.8\space m/s^{2})\times13\space m$
$V=\frac{\sqrt {254.8\space m^{2}/s^{2}}}{sin63^{\circ}}=17.9\space m/s$
