Question

A particle projected from the level ground just clears in it's ascent a wall $30m$ high and $120\sqrt{3}$ away measured horizontally. The time since projection to clear the wall is two second. It will strike the ground in the same horizontal plane from the wall on the other side at a distance of

• A. $150\sqrt{3}m$
• B. $180\sqrt{3}m$
• C. $120\sqrt{3}m$
• D. $210\sqrt{3}m$

Answer 1

Answer: (B)

$180\sqrt{3}m$

$S_y=ut-\frac{1}{2}g{t}^2\Rightarrow 30=\left(u\sin \theta \right)\left(2\right)-2g\Rightarrow u\sin \theta =25$
$S_x=u\cos \theta t\Rightarrow 2u\cos \theta =120\sqrt{3}$

Range $=\frac{u^2\sin 2\theta }{g}=\frac{\left(u\cos \theta \right)\left(u\sin \theta \right)2}{g}$
$=\frac{\left(60\sqrt{3}\right)\times 25\times 2}{10}=300\sqrt{3}$

Hence, distance from wall $=300\sqrt{3}-120\sqrt{3}$
$=180\sqrt{3}$ m