Question

A ball is projected from the ground at angle 0 with the horizontal. After 1 sec 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 ? (Take $g=10{ms}^{-2}$)

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

Answer 1

The correct option is C $10\sqrt{{5ms}^{-1}}$

Let the tangent make an angle $\alpha$ with the horizontal .

The value of g is given to be $10m{s}^{-2}$

For this motion under gravity

$\tan \alpha =\frac{usin\theta -gt}{ucos\theta }$ // where $u$ is the velocity .

Case 1 :

When $\alpha ={45}^{\circ }$ then $t=1sec$

$1=\frac{usin\theta -g}{ucos\theta }$ ///as $\tan {45}^{\circ }=1$

Or, $ucos\theta =usin\theta -g$ --------- (i)

Case 2 :

When the ball is moving horizontally ,

$\alpha =0^{\circ }$ then $t=2secs$

$0=\frac{usin\theta -g}{ucos\theta }$ //as $\tan 0=0$

Or, $usin\theta -2g=0$

$usin\theta =2g$ --------- (ii)

From (ii)

$ucos\theta =2g-g=g$

$ucos\theta =g$ ------ (iii)

Squaring and adding eq (ii) and (iii)

$u^2(si{n}^2\theta +co{s}^2\theta )=5g^2$

Or , $u=g\sqrt{5}=10\sqrt{5}m{s}^{-1}$