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 angle ${45}^{\circ }$ with horizontal. It is again projected from the same point at height $h$ 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. $ta{n}^{-1}\left(4\right)$
• B. $ta{n}^{-1}\left(5\right)$
• C. $ta{n}^{-1}\left(\sqrt{5}\right)$
• D. $ta{n}^{-1}\left(\sqrt{3}\right)$

Answer 1

Answer: (C)

$ta{n}^{-1}\left(\sqrt{5}\right)$

In first case, Let initial velocity be $u$ the initial velocity in horizontal direction is $ucos{30}^o$, in vertical direction is $usin{30}^o$.

As final angle when hitting ground is ${45}^o$, so final velocity in vertical direction is equal in magnitude to final velocity in horizontal direction, $ucos{30}^o$.

Using the formula, $v^2-u^2=2as$, in vertical direction,

$(-ucos{30}^o{)}^2-(usin{30}^o{)}^2=2gh$

$\Rightarrow u=2\sqrt{gh}$

In second case, Let final velocity in vertical direction be $v_1$. the initial velocity in horizontal direction is $ucos{60}^o$, in vertical direction is $usin{60}^o=2\sqrt{gh}\times 0.5=\sqrt{gh}$.

Using the formula, $v^2-u^2=2as$, in vertical direction,

$(v_1{)}^2-(usin{60}^o{)}^2=2gh$

$\Rightarrow v_1=\sqrt{5gh}$

So the final angle is, $ta{n}^{-1}\left(\frac{\sqrt{5gh}}{\sqrt{gh}}\right)=ta{n}^{-1}\sqrt{5}$