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 $\vec{u}_0 = u_0 \hat{x}$ and falls on the ground at a distance $d$ from the building making an angle $\theta$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_1$. The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $1/\sqrt{3}$. Which of the following statement(s) is(are) correct?

Step 1: Find the launch speed of the ball.

Since the slide is frictionless, mechanical energy is conserved as the ball descends a height $h$ to the terrace.

$$mgh = \frac{1}{2}m u_0^2$$ $$u_0 = \sqrt{2gh}$$

Thus, the launch velocity is $\vec{u}_0 = \sqrt{2gh}\,\hat{x}$.

Step 2: Find the time of flight and the horizontal distance $d$.

The ball is launched horizontally from the terrace at a height of $3h$ above the ground.

$$3h = \frac{1}{2}g t^2$$ $$t = \sqrt{\frac{6h}{g}}$$

The horizontal distance $d$ traveled before hitting the ground is:

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

Step 3: Find the angle of impact $\theta$ with the horizontal.

At the moment of impact, the horizontal and vertical components of the velocity are:

$$v_x = u_0 = \sqrt{2gh}$$ $$v_z = -gt = -\sqrt{6gh}$$

The angle $\theta$ with the horizontal is given by:

$$\tan\theta = \frac{|v_z|}{v_x}$$ $$\tan\theta = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3}$$ $$\theta = 60^\circ$$

Step 4: Find the velocity $\vec{v}$ after the bounce.

The coefficient of restitution of the ground is $e = \frac{1}{\sqrt{3}}$. The horizontal component of velocity remains unchanged, while the vertical component is reversed and scaled by $e$.

$$v_x' = v_x = \sqrt{2gh}$$ $$v_z' = e|v_z| = \frac{1}{\sqrt{3}}\sqrt{6gh} = \sqrt{2gh}$$

The velocity vector $\vec{v}$ immediately after the bounce is:

$$\vec{v} = \sqrt{2gh}\,\hat{x} + \sqrt{2gh}\,\hat{z}$$ $$\vec{v} = \sqrt{2gh}(\hat{x} + \hat{z})$$

Step 5: Find the maximum height $h_1$ reached after the bounce and the ratio $d/h_1$.

The maximum height $h_1$ depends on the vertical component of the bounce velocity.

$$h_1 = \frac{(v_z')^2}{2g}$$ $$h_1 = \frac{2gh}{2g} = h$$

The ratio of the horizontal distance $d$ to the rebound height $h_1$ is:

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

Answer: The correct statements are:

• The launch speed is $u_0 = \sqrt{2gh}$.
• The impact angle is $\theta = 60^\circ$.
• The ratio of the distance to the rebound height is $\dfrac{d}{h_1} = 2\sqrt{3}$.