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 $2\, s$ 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
$\tan \alpha=\frac{v_{y}}{v_{x}}=\frac{u \sin \theta-g t}{u \cos \theta}$
Given : $\alpha=45^{\circ}$, when $t=1 s$
$\alpha=0^{\circ}$, when $t=2 s$
This gives $u \cos \theta=u \sin \theta-g$
and $u \sin \theta-2 g=0$
Solving we have, $u \sin \theta=2 g$
$ u \cos \theta=g $
$\therefore u=\sqrt{5} \,g=10 \sqrt{5} m / s$