JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Kinematics

The time of flight depends only on the vertical component of the initial velocity. For the first ball thrown vertically upward with velocity $u$, time of flight is $T=\frac{2u}{g}$.

For the second ball thrown at angle $\theta$ with the vertical, the vertical component of velocity is $u\cos \theta$. For the same time of flight:

$\frac{2u\cos \theta }{g}=\frac{2u}{g}$

This implies $\cos \theta =1$, meaning $\theta =0$ (thrown vertically).

The maximum height is given by $H=\frac{u_{\text{vertical}}^2}{2g}$.

For both balls: $H_1=\frac{u^2}{2g}$ and $H_2=\frac{(u\cos \theta {)}^2}{2g}=\frac{u^2}{2g}$

Therefore, $\frac{H_1}{H_2}=1=\frac{1}{x}$, giving $x=1$.