Question

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

Answer 1

Answer:

$\displaystyle \theta=\tan^{-1}(4)$

Explanation:

Let projectile angle be θ .

Given :

Range ( R ) = Maximum Height ( H )

We have formula for Range

$\displaystyle R=\frac{u^2\sin2\theta}{g} \ ..(i)$

We also have for Height

$\displaystyle H_{max}=\frac{u^2\sin^2\theta}{2g} \ ..(ii)$

Given ( i )  and ( ii ) are equal

$\displaystyle R=H_{max}\\\\\displaystyle \frac{u^2\sin2\theta}{g}=\frac{u^2\sin^2\theta}{2g}\\\\\displaystyle \frac{u^2\sin2\theta}{1}=\frac{u^2\sin^2\theta}{2}\\\\\displaystyle \frac{\sin2\theta}{1}=\frac{\sin^2\theta}{2}\\\\\displaystyle 2=\frac{\sin^2\theta}{\sin2\theta}\\\\\displaystyle \text{We know that \sin2\theta=2\sin\theta.\cos\theta }\\\\\displaystyle \text{Rewrite \sin^2\theta as \sin\theta.\sin\theta }$

$\displaystyle 2=\frac{\sin\theta.\sin\theta}{2\sin\theta.\cos\theta}\\\\\displaystyle 2=\frac{\sin\theta }{2\cos\theta}\\\\\displaystyle 4=\frac{\sin\theta }{\cos\theta}\\\\\displaystyle \tan\theta=4\\\\\displaystyle \theta=\tan^{-1}(4)$

Thus we get answer.