JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Kinematics

The ball starts from rest at height $h$ above the terrace and slides down frictionlessly. By energy conservation, when it leaves the slide horizontally: $mgh=\frac{1}{2}m{u}_0^2$, giving $u_0=\sqrt{2gh}$, so option (A) is correct.

The ball then falls from height $3h$ with initial horizontal velocity $u_0=\sqrt{2gh}$. Just before hitting the ground, the vertical component is $v_y=\sqrt{2g(3h)}=\sqrt{6gh}$ downward, while horizontal component remains $v_x=\sqrt{2gh}$.

After bouncing with coefficient of restitution $e=1/\sqrt{3}$, the vertical component becomes $v_y^{\prime }=e\cdot \sqrt{6gh}=\frac{1}{\sqrt{3}}\cdot \sqrt{6gh}=\sqrt{2gh}$ upward, while horizontal component is unchanged. Thus $\vec{v}=\sqrt{2gh}\hat{x}+\sqrt{2gh}\hat{z}$ (taking $\hat{z}$ upward). Option (B) has a minus sign, so it's incorrect.

The angle with horizontal: $\tan \theta =\frac{\sqrt{6gh}}{\sqrt{2gh}}=\sqrt{3}$, giving $\theta =60°$. Option (C) is correct.

Maximum height after bounce: $h_1=\frac{(v_y^{\prime }{)}^2}{2g}=\frac{2gh}{2g}=h$.

Horizontal distance: $d=v_x\cdot t$ where $t=\sqrt{\frac{2(3h)}{g}}$ is the fall time. Thus $d=\sqrt{2gh}\cdot \sqrt{\frac{6h}{g}}=2\sqrt{3}h$.

Therefore $d/h_1=2\sqrt{3}h/h=2\sqrt{3}$. Option (D) is correct.