Why do projectiles with angle 45 make the biggest range?

why do projectiles with angle 45 make the biggest range?

Mar 13, 2018

If a projectile is thrown with a velocity $u$ with an angle of projection $\theta$,its range is given by the formula, $R = \frac{{u}^{2} \sin 2 \theta}{g}$

Now,if $u$ and $g$ are fixed, $R \propto \sin 2 \theta$

So,$R$ will be maximum when $\sin 2 \theta$ will be maximum.

Now,maximum value of $\sin 2 \theta$ is $1$

if, $\sin 2 \theta = 1$

so,$\sin 2 \theta = \sin 90$

so,$2 \theta = 90$

or, $\theta = {45}^{\circ}$

That means,when angle of projection is ${45}^{\circ}$ range is the maximum.