### Kinematics

Find the time in terms of the y component of the velocity at the beginning:
$v_{y} = v_{0y}-gt\\ 0 = v_{0y}-gt\\ t = \frac{v_{0y}}{g}$
Now use this time to find the highest point:\begin{eqnarray*}
$y = v_{0y}t-\frac{1}{2}gt^{2}\\ y_{max} = v_{0y}\left(\frac{v_{0y}}{g}\right)-\frac{1}{2}g\left(\frac{v_{0y}}{g}\right)^{2}\\ = \frac{v_{0y}^{2}}{2g}.$
We know $v_{0y}$ because it is just the part of the velocity projected
on the y axis and so:
$y_{max}=\frac{(v_{0}sin\varphi_{0})}{2g}.$