a) Projectile motion
Initial velocity of the projectile object = v_0
Angle between the horizontal(x-axis) and velocity = \theta

Initial velocity components:
Velocity along the x axis \vec{v_{0x}} = v_0 \cos \theta
Velocity along the y axis \vec{v_{0y}} = v_0 \sin \theta
Velocity after t seconds at P
Horizontal component v_x
Vertical components v_y
Using the equations motion to find the velocity components
First motion equation:
v_y = v_{0y} - gt
v_y = - gt
v_x = v_{0x}
The angle between the x-axis and the velocity
\tan \theta = \frac{v_y}{v_x}
\theta = \tan^{-1} (\frac{v_{0y} - gt}{v_{0x}})
Hence, proved
(b)
The maximum projectile height h_m=\frac{(v_0\sin\theta_0)^2}{2g}............................(1)
Horizontal range of a projectile R = \frac{(v_0^2 \sin 2\theta_0)}{g} .............................(2)
Dividing equation (1) by equation (2)
\frac{Eq(1)}{Eq(2)}
\frac{h_m}{R} = \frac{(v_0 \sin \theta_0)^2}{2g} * \frac{g}{v_0^2 \sin 2 \theta_0}
\frac{h_m}{R} = \frac{v_0^2 \sin^2 \theta_0}{2} * \frac{1}{2 v_0^2 \sin \theta \cos \theta}
\frac{h_m}{R} = \frac{\sin \theta_0}{4\cos \theta}
\frac{4h_m}{R} = \frac{\sin \theta_0}{4\cos \theta}
\frac{4h_m}{R} = \tan \theta
\theta = \tan^{-1} \frac{4h_m}{R}
Hence, projection Angle, \theta = \tan^{-1} \frac{4h_m}{R}