A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3ℎ from the ground, as shown in the figure. A spherical ball of mass 𝑚 is released on the slide from rest at a height ℎ from the top of the terrace. The ball leaves the slide with a velocity 𝑢⃗ 0 = 𝑢0𝑥̂ and falls on the ground at a distance 𝑑 from the building making an angle 𝜃 with the horizontal. It bounces off with a velocity v⃗ and reaches a maximum height ℎ1. The acceleration due to gravity is 𝑔 and the coefficient of restitution of the ground is 1⁄√3. Which of the following statement(s) is(are) correct? (A) u⃗ 0 = √2𝑔ℎ𝑥̂ (B) v⃗ = √2𝑔ℎ(𝑥̂ − 𝑧̂) (C) 𝜃 = 60° (D) 𝑑/ℎ1 = 2√3

 The problem involves a spherical ball released from a height 

$h$ above the terrace of a building, sliding down a frictionless curved surface, and undergoing projectile motion before bouncing off the ground. The key steps are as follows:

1. Velocity when the ball leaves the slide ( $u_0$ ):

   • Using conservation of energy, the potential energy at height $h$ is converted to kinetic energy.

   • $mgh=\frac{1}{2}m{u}_0^2$

   • Solving for $u_0$, we get $u_0=\sqrt{2gh}$

   • Statement (A) is correct.

2. Projectile motion and angle $\theta$:

   • The ball is projected horizontally from a height $3h$. Time to fall is $t=\sqrt{\frac{6h}{g}}$.

   • Horizontal distance $d=u_0\cdot t=\sqrt{2gh}\cdot \sqrt{\frac{6h}{g}}=2h\sqrt{3}$.

   • Vertical velocity just before impact $v_y=gt=\sqrt{6gh}$.

   • Angle $\theta$ with the horizontal: $\tan \theta =\frac{v_y}{u_0}=\sqrt{3}\Rightarrow \theta =60^{\circ }$.

   • Statement (C) is correct.

3. Velocity after bouncing ( $\vec{v}$ ) and maximum height $h_1$:

   • Coefficient of restitution $e=\frac{1}{\sqrt{3}}$.

   • Vertical velocity after collision: $v_y^{\mathrm{\prime }}=e\cdot v_y=\frac{1}{\sqrt{3}}\cdot \sqrt{6gh}=\sqrt{2gh}$.

   • Horizontal velocity remains $u_0=\sqrt{2gh}$.

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

   • Ratio $\frac{d}{h_1}=\frac{2h\sqrt{3}}{h}=2\sqrt{3}$.

   • Statement (D) is correct.

4. Velocity after bouncing ( $\vec{v}$ ):

   • The velocity after bouncing has horizontal component $\sqrt{2gh}$ and vertical component $\sqrt{2gh}$.

   • The correct velocity should be $\sqrt{2gh}(\widehat{x}+\widehat{z})$, not $\sqrt{2gh}(\widehat{x}-\widehat{z})$.

   • Statement (B) is incorrect.

Final Answer

\boxed{ACD}