Question

Prove that the maximum horizontal range of four times the maximum height attained by a projectile which is fired along the required oblique direction.

Answer 1

We know that,

The range of the projectile $R=\frac{u^{2}sin2\theta }{g}$,R will be the maximum, if $sin2\theta =1\Rightarrow 2\theta ={90}^0\Rightarrow \theta ={45}^0$

Then,

$R_{max}=\frac{u^2}{g}$

So the maximum height attained by the projectile is:

$H=\frac{u^{2}si{n}^2{45}^0}{2g}=\frac{u^2}{g}\times \frac{1}{2}=\frac{R_{max}}{4}\Rightarrow R_{max}=4H$