Question

At what angle should a projectile be thrown such that the horizontal range of the projectile will be equal to half of its maximum value?

Answer 1

We know: $R=\frac{u^2\sin 2\theta }{g}$ and $R_{max}=\frac{u^2}{g}$
Given $R=\frac{1}{2}{R}_{max}$
or $R=\frac{u^2\sin 2\theta }{g}=\frac{1}{2}{R}_{max}=R_{max}=\frac{u^2}{2g}$
$\Rightarrow \sin 2\theta =\frac{1}{2}or2\theta ={30}^o\Rightarrow \theta ={15}^o$
Also $R=\frac{u^2\sin (\pi -2\theta )}{g}$
Now, $\sin (\pi -2\theta )=\frac{1}{2}\Rightarrow {180}^o-2\theta ={30}^o\Rightarrow \theta ={75}^o$