Answer to Question #120140 in Mechanics | Relativity for Arif

Question

A stone is thrown from the level ground (as in projectile motion). It is observed that the maximum height of
its flight is equal to its horizontal range. Find the launch angle θ0.

Answer 1

Let, $H$ is the maximum height,$R$ is the horizontal range, $u$ is initial velocity and $\theta$ is the projectile angle.

Now, we have given $H=R$ , we have to find $\theta$ .

Since, we know that

$H=\frac{u^2\sin^2(\theta)}{2g}$

And

$R=\frac{u^2\sin(2\theta)}{g}$

Thus,

$H=R\implies \sin^2(\theta)=2\sin(2\theta)\\ \implies \sin^2(\theta)=4\sin(\theta)\cos(\theta)\\ \implies \tan(\theta)=4\implies\theta=\tan^{-1}(4)\approx 76^{\circ}$