Question

A particle at a height '$h$' from the ground is projected with an angle ${30}^{\circ }$ from the horizontal. It strikes the ground making angles ${45}^{\circ }$ with horizontal. It is again projected from the same point with the same speed but with an angle of ${60}^{\circ }$ with horizontal. Find the angle it makes with the horizontal when it strikes the ground.

• A. ${\tan }^{-1}\left(4\right)$
• B. ${\tan }^{-1}\left(5\right)$
• C. ${\tan }^{-1}\left(\sqrt{5}\right)$
• D. ${\tan }^{-1}\left(\sqrt{3}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\sqrt{5}\right)$

In first case, Let initial velocity be $u$. The initial velocity in horizontal direction is $u\cos {30}^{\circ }$, in vertical direction is $u\sin {30}^{\circ }$.
As final angle, when hitting ground is ${45}^{\circ }$, so final velocity in vertical direction is equal in magnitude to final velocity in horizontal direction, $u\cos {30}^{\circ }$.
Using the formula, $v^2–u^2=2as$, in vertical direction,
$(–u\cos {30}^{\circ }{)}^2–(u\sin {30}^{\circ }{)}^2=2gh$
$\Rightarrow u=2\sqrt{gh}$
In second case, Let final velocity in vertical direction be $v_1$. The initial velocity in horizontal direction is $u\cos {60}^{\circ }$=$2\sqrt{gh}\times 0.5=\sqrt{gh}$,
in vertical direction is $u\sin {60}^{\circ }=2\sqrt{gh}\times \frac{\sqrt{3}}{2}=\sqrt{3gh}$.
Using the formula $v^2–u^2=2as$, in vertical direction.
$(v_1{)}^2-(u\sin {60}^{\circ }{)}^2=2gh$
$\Rightarrow v_1=\sqrt{5gh}$
So, the final angle is ${\tan }^{-1}\left(\frac{\sqrt{5gh}}{\sqrt{gh}}\right)={\tan }^{-1}\sqrt{5}$