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(A-D) $1$. Velocity at the bottom of the slide: Using conservation of energy,$\frac{1}{2} m u_0^2 = mgh$,so $u_0 = \sqrt{2gh}$. Thus,$\vec{u}_0 = \sqr

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$A, C, D$

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(A-D) $1$. Velocity at the bottom of the slide: Using conservation of energy,$\frac{1}{2} m u_0^2 = mgh$,so $u_0 = \sqrt{2gh}$. Thus,$\vec{u}_0 = \sqrt{2gh} \hat{x}$. Statement $(A)$ is correct.$2$. Motion of the ball: The ball undergoes projectile motion from height $H = 3h$. The time of flight $T = \sqrt{2H/g} = \sqrt{6h/g}$.The horizontal velocity $v_x = u_0 = \sqrt{2gh}$. The vertical velocity just before impact is $v_{1y} = \sqrt{2gH} = \sqrt{2g(3h)} = \sqrt{6gh}$.$3$. Impact with the ground: The angle $	heta$ with the horizontal is given by $	an 	heta = \frac{v_{1y}}{v_x} = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3}$. Thus,$	heta = 60^{\circ}$. Statement $(C)$ is correct.$4$. Velocity after bounce: The horizontal component remains $v_x = \sqrt{2gh}$. The vertical component after bounce is $v_{2y} = e v_{1y} = \frac{1}{\sqrt{3}} \sqrt{6gh} = \sqrt{2gh}$. Thus,$\vec{v} = \sqrt{2gh} \hat{x} + \sqrt{2gh} \hat{z}$. Statement $(B)$ is correct.$5$. Maximum height $h_1$: $h_1 = \frac{v_{2y}^2}{2g} = \frac{2gh}{2g} = h$.$6$. Distance $d$: $d = v_x T = \sqrt{2gh} \cdot \sqrt{6h/g} = \sqrt{12h^2} = 2\sqrt{3}h$.$7$. Ratio $d/h_1 = \frac{2\sqrt{3}h}{h} = 2\sqrt{3}$. Statement $(D)$ is correct.

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