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 $4\, s$, the stone is moving at an angle $30^{\circ}$ to the horizontal and after another $2\, s$ it is travelling horizontally. The magnitude of the initial velocity of the stone is (Acceleration due to gravity, $\left.g=10\, ms ^{-2} .\right)$

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

Answer 1

Answer: (A)

Given, acceleration due to gravity, $g=10 \,m / s ^{2}$
After $4s$, angle between stone and horizontal plane, $\theta=30^{\circ}$
After $t=4 s$, 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, $t=2+4=6\, s$
After $t=6\, s$, equation of horizontal projectile motion, $\theta=0^{\circ}$
$\tan 0^{\circ}=\frac{v \sin \theta-g(6)}{v \times \cos 0} \,\,\,\begin{bmatrix}\because \cos 0^{\circ}=1 \\ \tan 0^{\circ}=0\end{bmatrix}$
or $v \sin \theta-g(6)=0%$
$v \sin \theta=60 \,\,\,\ldots(iii)$
$[\because$ Given, $g=10]$
From Eq. (ii), At $t=4\, s$, 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 $v=40 \sqrt{3}\, m / s$