Question

We know that for maximum horizontal range,
( theta=45^{0} )
and horizontal range, R=
( frac{u^{2}}{g} )
( therefore ) max imum height ( , h=frac{u^{2} sin ^{2} theta}{2 g}=frac{u^{2} sin ^{2} 45^{0}}{2 g} )
[
begin{aligned}
h &=frac{u^{2}}{4 g}
therefore h &=frac{R}{4}
end{aligned}
]

# 54. When the maximum range of a projectile is R, then its maximum height is (a) R (b)

Solution