Question

prove that the maximum horizontal range is four times the maximum height attained by the projectile when fired at an inclination so as to have maximum horizontal range ​

Answer 1

The maximum range is achieved when θ=π/4

R = v^2 sin2θ/g

h = v^2 sin^2 θ / 2g

R/h = 4cotθ

at θ=π/4, R/h = 4

Answer 2

To have maximum horizontal range or to have maximum distance covered, the object should be thrown at 45° angle from the x-axis if it's plotted in graph.

$range = \frac{ {u}^{2} }{g}$
${sin}^{2} 45 = \frac{1}{2}$

$height = \frac{ {u}^{2} {sin }^{2} \alpha }{2g}$

taking alpha as 45 we have
$height = \frac{ {u}^{2} }{4g}$
therefore,

Range = 1/4 Height....... (cancelling common terms

Therefore four times range is height mathematically.

So, the maximum horizontal range is four times the maximum height attained by the projectile when fired at an inclination so as to have maximum horizontal range.