### Kinematics

(a) The initial velocity up is $v_{0y}=v_{0}sin\phi_{0}.$The time
to reach the highest point is $t=v_{0y}/g$. The highest point is
$H=gt^{2}/2$. We combine all this:
$H=\frac{1}{2}g\left(\frac{v_{0}sin\phi_{0}}{g}\right)^{2}=\frac{v_{0}^{2}sin^{2}\phi_{0}}{2g}.$
The range is $R=(v_{o}^{2}/g)sin2\phi_{0}=2(v_{0}^{2}/g)sin\phi_{0}cos\phi_{0}$.
Since $tan\theta=H/(R/2)$, we have
$tan\theta=\frac{2H}{R}=\frac{v_{0}^{2}sin^{2}\phi_{0}/g}{2(v_{0}^{2}/g)sin\phi_{0}cos\phi_{0}}=\frac{1}{2}tan\phi_{0}.$
(b)$\theta=arctan(0.5tan45^{o})=26.6^{o}$.