Question

A projectile is fired in such a way that it's horizontal range is equal to three times it's maximum height.

What is the angle of projection?

Answer 1

Step-1: Formula used:

The expression for the time of flight, into the formula for $x$ distance to find the range (denoted $d$) is
$\mathrm{d}=\mathrm{x}\left({\mathrm{t}}_{\mathrm{f}}\right)={\mathrm{v}}_{\circ }\mathrm{cos\theta }.{\mathrm{t}}_{\mathrm{f}}=\frac{2{{\mathrm{v}}^2}_{\circ }\mathrm{sin\theta cos\theta }}{\mathrm{g}}=\frac{{{\mathrm{v}}^2}_{\circ }\sin 2\mathrm{\theta }}{\mathrm{g}}$

Maximum height is given by

$\mathrm{h}=\frac{{{\mathrm{v}}^2}_{\circ }{\sin }^2\mathrm{\theta }}{2\mathrm{g}}$

$\mathrm{d}=3\mathrm{h}$,

By trigonometric identity

$\sin 2\mathrm{\theta }=2\mathrm{sin\theta cos\theta }$

Step-2: Calculating the angle of projection:

The angle of projection is determined by the following steps

We know that,

$\mathrm{d}=3\mathrm{h}$

$\frac{{{\mathrm{v}}^2}_{\circ }\sin 2\mathrm{\theta }}{\mathrm{g}}=\frac{{3{\mathrm{v}}^2}_{\circ }{\sin }^2\theta }{2\mathrm{g}}$

$\sin 2\mathrm{\theta }=\frac{3{\sin }^2\mathrm{\theta }}{2}$

Where $\sin 2\mathrm{\theta }=2\mathrm{sin\theta cos\theta }$

$2\mathrm{sin\theta cos\theta }=\frac{3}{2}{\sin }^2\mathrm{\theta }$

$\mathrm{sin\theta }(\frac{3}{2}\mathrm{sin\theta }-2\mathrm{cos\theta })=0$
$\mathrm{sin\theta }=0$or $\mathrm{tan\theta }=\frac{4}{3}$

$\mathrm{\theta }={\tan }^{-1}\left(\frac{4}{3}\right)$

$\mathrm{\theta }=53.1^{\circ }$

Therefore the angle of projection is $53.1^{\circ }$.