Grigori
0

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}