Question

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is:
$\begin{align}
  & \text{A}.{{\tan }^{-1}}\dfrac{H}{R} \\
 & \text{B}.{{\tan }^{-1}}\dfrac{2H}{R} \\
 & \text{C}.{{\tan }^{-1}}\dfrac{4H}{R} \\
 & \text{D}.{{\tan }^{-1}}\dfrac{4R}{H} \\
\end{align}$

Answer 1

Hint: R, the horizontal range of a projectile is the total horizontal distance that travels in a motion. And H is the maximum height a body attains in a projectile before it starts to fall downwards. Here we are asked to find the angle of projection with the horizontal, to solve for that we divide maximum height (H) by range (R).

Formula used: $H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ And $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$

Complete step by step answer:
Consider a projectile motion as shown in the above figure. The initial velocity of the projectile is $um/s$, H is the maximum height of the projectile and R is its horizontal range.
We know that maximum height of the projectile H is given by the equation,
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
And horizontal range is given by the equation,
$R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$
In the question, we are asked to find the angle of projection, i.e. $\theta $.
To find $\theta $ , let us divide the maximum height of the projectile by horizontal range.
Thus we get,
$\dfrac{H}{R}=\left( \dfrac{\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}}{\dfrac{{{u}^{2}}\sin 2\theta }{g}} \right)$
$\dfrac{H}{R}=\left( \dfrac{{{u}^{2}}{{\sin }^{2}}\theta \times g}{{{u}^{2}}\sin 2\theta \times 2g} \right)$
By solving this equation, we get
$\dfrac{H}{R}=\dfrac{{{\sin }^{2}}\theta }{2\sin 2\theta }$
 In the above equation, ${{\sin }^{2}}\theta $ can be written as $\sin \theta \times \sin \theta $
And $\sin 2\theta $ can be written as $2\sin \theta \cos \theta $.
Hence the equation can be rewritten as
$\dfrac{H}{R}=\dfrac{\sin \theta \times \sin \theta }{2\times 2\times \sin \theta \times \cos \theta }$
$\dfrac{H}{R}=\dfrac{\sin \theta }{4\cos \theta }$
We know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $. Therefore the equation can be written as
$\dfrac{H}{R}=\dfrac{1}{4}\tan \theta $
From this equation we get,
$\tan \theta =4\dfrac{H}{R}$
We have to find $\theta $, therefore
$\theta ={{\tan }^{-1}}\dfrac{4H}{R}$
Therefore the angle of projection of the projectile, $\theta ={{\tan }^{-1}}\dfrac{4H}{R}$.

So, the correct answer is “Option D”.

Note: Projectile motion is a motion that a body experiences when it is thrown or projected near the surface of earth. In such a motion the projectile moves in a curved path only under the influence of gravity. Hence the path of a projectile will always be a parabola.