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 $g=10m{s}^{-2}$ )

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

Answer 1

Answer: (D)

Let $V_1$ be the initial speed at '$O$' and ${}^{\prime }{\theta }^{\prime }$ be the initial angle of projection

After two seconds particle reaches at $Q$ and $\theta =30^{\circ }$
Hence, $V_2\cos 30^{\circ }=V_1\cos \theta$
$V_2\cdot \frac{\sqrt{3}}{2}=V_1\cos \theta$
$V_2=\frac{2}{\sqrt{3}}{V}_1\cos \theta \dots \left(i\right)$
By equation of motion, when particle reaches from $O$ to $Q$
$V_2\sin 30^{\circ }=V_1\sin \theta -2g$
$\frac{V_2}{2}=V_1\sin \theta -2g$
$\frac{1}{2}\cdot \frac{2}{\sqrt{3}}{V}_1\cos \theta =V_1\sin \theta -2g$
$V_1\cos \theta =\sqrt{3}{v}_1\sin \theta -2\sqrt{3}g\dots \left(ii\right)$
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 $v_1=\frac{3g}{\sin \theta }\dots \left(iii\right)$
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 $\left(iii\right)$,
$V_1=\frac{3g}{\sin 60^{\circ }}=\frac{3g}{\frac{\sqrt{3}}{2}}$
$=2\sqrt{3}g$
$=2\sqrt{3}\times 10$
$=20\sqrt{3}m/s$