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}.