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)

$10\sqrt{5}\,m{{s}^{-1}}$

Answer 2

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, $u \sin \theta=2 g$ $u \cos \theta=g$ $\therefore u=\sqrt{5} \,g=10 \sqrt{5} m / s$