Question

$AB$ is an inclined plane of inclination $30^{\circ }$ with horizontal. Point $O$ is $20m$ above point $A$. A particle is projected horizontally from $O$ leftwards, and it collides with the plane $AB$ perpendicularly. Speed of the particle at the time of projection should be $\left(g=10m/s^2\right)$

• A. $13m/s$
• B. $8\sqrt{3}m/s$
• C. $4\sqrt{5}m/s$
• D. $2\sqrt{3}m/s$

Answer 1

Answer: (C)

In $x$ direction: $v_I=u_I+a_{I}t$
$\Rightarrow 0=u\cos 30^{\circ }-g\sin 30^{\circ }t$
$\Rightarrow u\sqrt{3}=gt$ .... (i)

In $y$ direction: $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 20\cos 30^{\circ }=u\cos 60^{0}t+\frac{1}{2}g\cos 30^{\circ }{t}^2$
$\Rightarrow 20\sqrt{3}=ut+\frac{1}{2}g{t}^2\sqrt{3}$
$\Rightarrow 20\sqrt{3}=\frac{u^2\sqrt{3}}{g}+\frac{\sqrt{3}}{2}g\cdot \frac{3u^2}{g^2}\Rightarrow 20=\frac{5u^2}{2g}$
$\Rightarrow u=\sqrt{80}m/s=4\sqrt{5}m/s$