Question

A ball is projected from the ground at angle $\theta$ with the horizontal. After $1s$, 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

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

x and y components are as follows;-

$x=u\cos \theta t$ and $vx=u\cos \theta$

$y=u\sin \theta t-\frac{1}{2}g{t}^2⟶1$

$v_y=u\sin \theta -gt⟶2$

The horizontal component of speed $v_x$ is always constant. So the speed is minimum when the vertical component $v_y=0$

Putting this in equation $2$

$u\sin \theta =2g$

The ball is moving at angle $45°$ direction of velocity of the project at time $t=1s$

Then $\tan 45°=1=\frac{v_y}{v_x}=\frac{u\sin \theta -g\times 1}{u\cos \theta }$

$\Rightarrow u\cos \theta =g$

$\tan \theta =2$

$\theta ={\tan }^{-1}2$

$\sin \theta =\frac{2}{\sqrt{5}}$

$\cos \theta =\frac{1}{\sqrt{5}}$

and $u=\sqrt{5}g$

$=\sqrt{5}\times 10m/s$

$u=10\sqrt{5}m/s$