Question

show that the horizontal range is maximum when angle of projectile is 45°​

Answer 1

The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. ... That means that the best way to launch a high-altitude projectile is to send it flying at a 90-degree angle to the ground—straight up.

Answer 2

we have to show that the horizontal range is maximum when angle of projectile is 45°.

Let a body is projected with speed u m/s at an angle $\theta$.

so, horizontal range , H = $\frac{u^2sin2\theta}{g}$

here it is clear that, horizontal range will be maximum only when $sin2\theta$ = 1

$sin2\theta=sin90^{\circ}$

$2\theta=90^{\circ}$

$\theta=45^{\circ}$

hence, horizontal range is maximum when angle of projectile is 45°