Question

Two seconds after projection, a projectile is moving at $ 30^{\circ} $ above the horizontal, after one more second, it is moving horizontally. The initial speed of the projectile is (Take $ g = 10\, ms^{-2} $ )

• A. $ 10 \, m/s $
• B. $ 10 \sqrt{3} \,m/s $
• C. $ 20 \, m/s $
• D. $ 20 \sqrt{3} \,m/s $

Answer 1

Answer: (D)

Let $V_{1}$ be the initial speed at '$O$' and $'\theta'$ be the initial angle of projection

After two seconds particle reaches at $Q$ and $\theta=30^{\circ}$
Hence, $V_{2}\,\cos \, 30^{\circ} = V_{1}\, \cos\, \theta$
$V_{2} \cdot \frac{\sqrt{3}}{2}=V_{1}\, \cos\, \theta $
$V_{2}=\frac{2}{\sqrt{3}}V_{1}\, \cos\, \theta \ldots\left(i\right)$
By equation of motion, when particle reaches from $O$ to $Q$
$V_{2}\, \sin \,30^{\circ}=V_{1}\,\sin\, \theta-2g$
$\frac{V_{2}}{2}=V_{1}\,\sin \, \theta-2g$
$\frac{1}{2}\cdot\frac{2}{\sqrt{3}}V_{1}\, \cos\, \theta =V_{1}\,\sin\,\theta-2g$
$V_{1}\,\cos\,\theta=\sqrt{3}v_{1}\, \sin \, \theta-2\sqrt{3}g \ldots\left(ii\right)$
After $3$ seconds, i.e. at top position, particle moves in horizontal direction, hence vertical component is zero
$\therefore O=V_{1}\, \sin \,\theta-g\times t $
$O=V_{1}\,\sin\,\theta-3g$
or $ v_{1}=\frac{3g}{\sin\,\theta} \ldots\left(iii\right)$
Putting the value of $V_{1}$ from Eq (iii) to Eq (ii), we have,
$3g \frac{\cos\,\theta}{\sin \,\theta}=\sqrt{3} \frac{3g}{\sin\,\theta} \cdot \sin\, \theta-2\sqrt{3}g$
$3g\, \cot \,\theta=3\sqrt{3}g-2\sqrt{3}g$
$3g\, \cot\,\theta=\sqrt{3}g$
$\cot \,\theta=\frac{\sqrt{3}g}{3g}$
$\cot\,\theta=\frac{1}{\sqrt{3}}$
$\cot \,\theta=\cot\,60^{\circ}$
$\therefore \theta=60^{\circ}$
$\therefore $ From Eq $\left(iii\right)$,
$V_{1}=\frac{3g}{\sin\,60^{\circ}}=\frac{3g}{\frac{\sqrt{3}}{2}}$
$=2\sqrt{3}g$
$=2\sqrt{3}\times10$
$=20\sqrt{3} \,m/ s$