Q 11-03-064JEE MainJEE Main 2024 (5 Apr, Shift 2)Easy
The maximum height reached by a projectile is $64\ \text{m}$. If the initial velocity is halved, the new maximum height of the projectile is ______ m.
Numerical value type. Enter your answer.
Answer: 16
$H = \dfrac{u^2\sin^2\theta}{2g} \propto u^2$. Halving $u$ divides $H$ by 4: $H' = \dfrac{64}{4} = 16\ \text{m}$.
Solution by Sreeraj P, M.Sc Physics