Question

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Answer 1

Hint
To solve this question, we have to use the formulae of the horizontal range and the maximum height for a projectile. Equating the two expressions will give out the desired result.
Formula Used: The formulae used in solving this question are
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where $R$ and $H$ are the respective horizontal range and the maximum height covered by a projectile thrown with an initial velocity $u$ at an angle of $\theta $ with the horizontal.

Complete step by step answer
Let the projectile be launched with an initial velocity $u$ at an angle of $\theta$ with the horizontal.
We know that the horizontal range covered by a projectile is given by
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ (1)
Also, we know that the maximum height reached by the projectile is given by
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ (2)
According to the question, the horizontal range and the maximum height covered by the projectile are equal. Therefore, on equating (1) and (2), we have
$\dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Cancelling ${u^2}$ from both the sides, we have
$\dfrac{{\sin 2\theta }}{g} = \dfrac{{{{\sin }^2}\theta }}{{2g}}$
Multiplying with $2g$ on both the sides
$\sin 2\theta = \dfrac{{{{\sin }^2}\theta }}{2}$
We know from trigonometry that $\sin 2\theta = 2\sin \theta \cos \theta $
Making this substitution in the above equation, we have
$2\sin \theta \cos \theta = \dfrac{{{{\sin }^2}\theta }}{2}$
Multiplying both sides by $2$
$4\sin \theta \cos \theta = {\sin ^2}\theta $
Subtracting $4\sin \theta \cos \theta = {\sin ^2}\theta $ from both the sides
${\sin ^2}\theta - 4\sin \theta \cos \theta = 0$
Taking $\sin \theta $ common, we have
$\sin \theta \left( {\sin \theta - 4\cos \theta } \right) = 0$
From here, we have two conditions
$\sin \theta = 0$ or (3)
$\sin \theta - 4\cos \theta = 0$
Or $\sin \theta = 4\cos \theta $
Dividing both sides by $\cos \theta $
$\tan \theta = 4$ (4)
From (3) we have
$\sin \theta = 0$
$ \Rightarrow \theta = {0^ \circ }$
This is not the case of the projectile motion. So, we reject this value of angle.
From (4) we have
$\tan \theta = 4$
$ \Rightarrow \theta = {\tan ^{ - 1}}4$
Hence, the angle of projection for which the horizontal range and maximum height are equal is equal to ${\tan ^{ - 1}}4$.

Note
Do not ignore a value of the angle of projection out of the two values of the angles which are found out after solving the trigonometric equation. In this case, the first value $\theta = {0^ \circ }$ was of no use, but it is not always true.