Question

A ball is projected from the ground at angle $\theta$ with the horizontal. After Is it is moving at angle $45^{\circ }$ with the horizontal and after $2s$ it is moving horizontally. What is the velocity of projection of the ball?

• A. $10\sqrt{3}m{s}^{-1}$
• B. $20\sqrt{3}m{s}^{-1}$
• C. $10\sqrt{5}m{s}^{-1}$
• D. $20\sqrt{2}m{s}^{-1}$

Answer 1

Answer: (C)

Suppose the angle made by the instantaneous velocity with the horizontal be $\alpha$. Then
$\tan \alpha =\frac{v_y}{v_x}=\frac{u\sin \theta -gt}{u\cos \theta }$
Given : $\alpha =45^{\circ }$, when $t=1s$
$\alpha =0^{\circ }$, when $t=2s$
This gives $u\cos \theta =u\sin \theta -g$
and $u\sin \theta -2g=0$
Solving we have, $u\sin \theta =2g$
$u\cos \theta =g$
$\therefore u=\sqrt{5}g=10\sqrt{5}m/s$