Question

A girl throws a ball with initial velocity v at an inclination of $45$ Find height and Range ?

Answer 1

Given velocity of the ball $v_0=v$ and angle of projectile $\theta ={45}^0$

We know time of flight $T=\frac{v_0\sin \theta }{g}=\frac{v}{10\sqrt{2}}$

as $\sin {45}^0=\frac{1}{\sqrt{2}}$ and g=10$m/s^2$

Maximum height$H=\frac{v^2{\sin }^2\theta }{2g}=\frac{v}{40}$ and

Horizontal range $R=\frac{v^2\sin 2\theta }{g}=\frac{v^2}{10}$