Question

Two boys conducted experiments on the projectile motion with stopwatch and noted some readings. As one boy throws a stone in air at the same angle with the horizontal, the other boy observes that after $4s$, the stone is moving at an angle $30^{\circ }$ to the horizontal and after another $2s$ it is travelling horizontally. The magnitude of the initial velocity of the stone is (Acceleration due to gravity, $g=10m{s}^{-2}.)$

• A. $40\sqrt{3}m{s}^{-1}$
• B. $20\sqrt{3}m{s}^{-1}$
• C. $10\sqrt{3}m{s}^{-1}$
• D. $50\sqrt{3}m{s}^{-1}$

Answer 1

Answer: (A)

Given, acceleration due to gravity, $g=10m/s^2$
After $4s$, angle between stone and horizontal plane, $\theta =30^{\circ }$
After $t=4s$, equation of the vertical projectile motion, when $\theta =30^{\circ }$
$\therefore \tan \theta =\frac{v\sin \theta -g(t)}{v\cos \theta }...(i)$
$\tan 30^{\circ }=\frac{v\sin \theta -g(4)}{v\cos \theta }...(ii)$
Total time to reach the stone at horizontal surface, $t=2+4=6s$
After $t=6s$, equation of horizontal projectile motion, $\theta =0^{\circ }$
$\tan 0^{\circ }=\frac{v\sin \theta -g(6)}{v\times \cos 0}\left[\begin{array}{c}\because \cos 0^{\circ }=1\\ \tan 0^{\circ }=0\end{array}\right]$
or $v\sin \theta -g(6)=0$
$v\sin \theta =60\dots (iii)$
$[\because$ Given, $g=10]$
From Eq. (ii), At $t=4s$, when particles travelling in horizontal direction,
$\therefore v\cos \theta =20\sqrt{3}...(iv)$
Now, magnitude of initial velocity,
$v=\sqrt{(v\sin \theta {)}^2+(V\cos \theta {)}^2}$
or [From Eqs. (iii) and (iv) ]
or $v=\sqrt{(60{)}^2+(20\sqrt{3}{)}^2}$
So, the magnitude of initial velocity of stone is $v=40\sqrt{3}m/s$