Mindyard | 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 u 0 = u 0 i ^ 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 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 3 . Which of the following statement(s) is(are) correct? u0 = 2ghi^, v = 2gh(i^−j^), θ = 60°, dh1=23

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 $u_{0} = u_{0} \hat{i}$ 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 v and reaches a maximum height h₁. The acceleration due to gravity is g and the coefficient of restitution of the ground is $\frac{1}{\sqrt{3}}$ . Which of the following statement(s) is(are) correct?

Options

u₀ = $\sqrt{2 g h} \hat{i}$

v = $\sqrt{2 g h} (\hat{i} - \hat{j})$

θ = 60°

$\frac{d}{h_{1}} = 2 \sqrt{3}$

Correct Answer :

Option A

u₀ = $\sqrt{2 g h} \hat{i}$

Option C

θ = 60°

Option D

$\frac{d}{h_{1}} = 2 \sqrt{3}$

Solution :

The correct options are:
u₀ = $\sqrt{2 g h} \hat{i}$ ,
θ = 60°, and
$\frac{d}{h_{1}} = 2 \sqrt{3}$ .

Step 1: Determine the horizontal velocity of the ball on leaving the slide
Using conservation of energy for the ball sliding down the frictionless surface of height h:

$m g h = \frac{1}{2} m u_{0}^{2}$

Solving for $u_{0}$ :

$u_{0} = \sqrt{2 g h}$

Since it leaves horizontally in the positive x-direction:

${\vec{u}}_{0} = \sqrt{2 g h} \hat{i}$

Thus, the first option is correct.

Step 2: Motion of the ball from the terrace to the ground
The height of the building is 3h. The vertical motion under gravity yields the time of flight t:

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

The horizontal distance d travelled before striking the ground is:

$d = u_{0} t = \sqrt{2 g h} \cdot \sqrt{\frac{6 h}{g}} = \sqrt{12 h^{2}} = 2 \sqrt{3} h$

Step 3: Finding the striking velocity and angle θ
Just before hitting the ground, the velocity components are:

Horizontal component: $v_{x} = u_{0} = \sqrt{2 g h}$
Vertical component (downwards): $v_{y} = g t = g \sqrt{\frac{6 h}{g}} = \sqrt{6 g h}$

The angle θ made by the trajectory with the horizontal at the moment of impact is:

$\tan θ = \frac{v_{y}}{v_{x}} = \frac{\sqrt{6 g h}}{\sqrt{2 g h}} = \sqrt{3}$

Therefore:

$θ = 60 °$

Thus, the third option is correct.

Step 4: Motion after bouncing from the ground
The coefficient of restitution is $e = \frac{1}{\sqrt{3}}$ .
During the collision with the smooth ground, the horizontal velocity component remains unchanged, while the vertical component reverses direction and its magnitude becomes:

$v_{y}^{'} = e v_{y} = \frac{1}{\sqrt{3}} \sqrt{6 g h} = \sqrt{2 g h}$

The rebound velocity vector is given by:

$\vec{v} = v_{x} \hat{i} + v_{y}^{'} \hat{k} = \sqrt{2 g h} \hat{i} + \sqrt{2 g h} \hat{k} = \sqrt{2 g h} (\hat{i} + \hat{k})$

(Note: This makes option 2 incorrect because the vertical unit vector after rebound points upwards, not downwards).

Step 5: Calculate maximum height h₁ reached after collision
The maximum height reached after the bounce depends solely on the vertical rebound velocity component $v_{y}^{'}$ :

$h_{1} = \frac{{(v_{y}^{'})}^{2}}{2 g} = \frac{2 g h}{2 g} = h$