Question

During projectile motion if the maximum height is equals to the horizontal range, then the angle of projection with the horizontal is
$\left( A \right){\tan ^{ - 1}}\left( 1 \right)$
$\left( B \right){\tan ^{ - 1}}\left( 2 \right)$
$\left( C \right){\tan ^{ - 1}}\left( 3 \right)$
$\left( D \right){\tan ^{ - 1}}\left( 4 \right)$

Answer 1

There are different parameters related to projectile motion which we can make use of like time of flight, horizontal range and maximum height. We know the formula of each of them and here to solve this problem we first equate maximum height with horizontal range. From there we can find the angle of projection.

Complete step by step solution:
Projectile Motion:
It is the motion of an object thrown or projected into the air, subject to only acceleration of gravity and its path is called a trajectory.
We know the formula for Maximum height and horizontal range,
Now,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \ldots \ldots \left( 1 \right)$
Where,
Maximum height of the projection is equal to H.
The initial velocity at which the object is projected is u.
The angle of projection is equal to $\theta$.
The acceleration due to gravity is g
$R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g} \ldots \ldots \left( 2 \right)$
Where,
Horizontal range is equals to R
And all other terms are the same as the maximum height.
Now according to the question equation $\left( 1 \right)$ and $\left( 2 \right)$ are equals to each other.
Hence equating both we will get,
$H = R$
$\Rightarrow \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$
Cancelling the common terms we will get,
$\dfrac{{\sin \theta }}{2} = 2\cos \theta$
Rearranging the above equation we will get,
$\dfrac{{\sin \theta }}{{\cos \theta }} = 2 \times 2$
$\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = 4$
We know from the trigonometric properties, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Hence the above equation will become,
$\tan \theta = 4$
Taking tan to other side we will get,
$\theta = {\tan ^{ - 1}}\left( 4 \right)$
Therefore the correct option is $\left( D \right)$.

Note:
The horizontal range of the projectile motion is also written as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ but here as we have to find the value of angle of projection hence for easy simplification we used $2\sin \theta \cos \theta$ in place of $\sin 2\theta$ . The value of u, g and angle of projection will remain same for both maximum height and horizontal range as both of them are the parameter of same projectile motion.