A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on t

Q: Why didn't we use the total height $4h$ for the potential energy?

Because we analyzed the motion in phases. For the slide phase, the ball only descends a vertical distance of $h$. The remaining $3h$ is covered during the free fall phase.

Q: Why is the horizontal velocity constant during the free fall?

Once the ball leaves the slide, the only force acting on it is gravity, which pulls straight down. Since there is no horizontal force (ignoring air resistance), the horizontal acceleration is zero, and the horizontal velocity remains constant at $\sqrt{2gh}$.

Q: I don't understand how the coefficient of restitution works here.

The coefficient of restitution $e$ tells us how much vertical speed is retained after a bounce. It is the ratio of the speed of separation to the speed of approach. Since the ground doesn't move, the ball's rebound vertical speed is simply $e$ times its impact vertical speed.

Q: Why did Option B have a negative z-component?

Option B proposed $\vec{v} = \sqrt{2gh}(\hat{x} - \hat{z})$. The $-\hat{z}$ implies the ball is moving downwards after the bounce. But physically, a ball bounces *upwards* off the ground, so its vertical velocity must be in the $+\hat{z}$ direction.

Q: Could we have used the standard range formula $R = \frac{u^2 \sin 2\theta}{g}$?

No! That formula only works when a projectile is launched and lands at the exact same vertical level. Here, the ball is launched horizontally from a height of $3h$, so we must use the fundamental kinematic equations to find the time of flight first.

Question

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 $	heta$ 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?

Options:

(A) $\vec{u}_0 = \sqrt{2gh}\hat{x}$

(B) $\vec{v} = \sqrt{2gh}(\hat{x} - \hat{z})$

(C) $	heta = 60^\circ$

(D) $d/h_1 = 2\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the terrace of a building, $3h$ meters above the ground. You release a small spherical ball from rest down a frictionless curved slide of height $h$.

The journey of this ball is a beautiful three-act play: a smooth slide, a free fall, and a dramatic bounce. Let's break down the physics of each phase step-by-step.

### Phase 1: The Frictionless Slide

Our first goal is to find the velocity of the ball exactly when it leaves the slide. Because the slide is perfectly frictionless, we can confidently invoke the **Conservation of Mechanical Energy**.

The ball starts from rest, so its initial kinetic energy is zero. As it descends a vertical height $h$, it loses gravitational potential energy and gains kinetic energy.

$$ mgh = \frac{1}{2}mu_0^2 $$

Solving for the exit speed $u_0$, we get:

$$ u_0 = \sqrt{2gh} $$

The problem states that the slide becomes horizontal at its lower end. This is a crucial geometric constraint! It means the velocity vector is purely in the positive x-direction.

$$ \vec{u}_0 = \sqrt{2gh}\hat{x} $$

This perfectly matches Option (A). We've secured our first correct statement!

### Phase 2: The Free Fall

Once the ball leaves the slide, it becomes a projectile. It is launched horizontally from a height of $3h$. During this free fall, gravity is the only force acting on it.

The horizontal velocity remains completely unaffected by gravity:

$$ v_{x1} = u_0 = \sqrt{2gh} $$

In the vertical direction, the ball starts with zero initial velocity and accelerates downwards at $g$ over a distance of $3h$. Using the third equation of motion ($v^2 = u^2 + 2as$), we find the vertical velocity just before impact:

$$ v_{z1}^2 = 0^2 + 2(-g)(-3h) \implies v_{z1} = -\sqrt{6gh} $$

Now, we can find the angle $	heta$ the velocity vector makes with the horizontal just before hitting the ground:

$$ 	an	heta = \left| \frac{v_{z1}}{v_{x1}} ight| = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3} $$

This gives us $	heta = 60^\circ$, which means Option (C) is also correct.

### Phase 3: The Bounce

Here is where many students make a silly mistake. The ball hits the ground and bounces. The ground is horizontal and smooth, meaning there are no horizontal forces during the collision. Therefore, the horizontal velocity remains exactly the same:

$$ v_x = \sqrt{2gh} $$

However, the vertical velocity is altered by the collision. The coefficient of restitution $e = 1/\sqrt{3}$ dictates how much vertical speed is retained. The ball reverses its vertical direction (it bounces *up*), and its new vertical speed is $e$ times the impact speed:

$$ v_z = e |v_{z1}| = \left(\frac{1}{\sqrt{3}}ight) \sqrt{6gh} = \sqrt{2gh} $$

So, the velocity vector immediately after the bounce is:

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

Notice the positive $\hat{z}$ component! Option (B) suggests a negative $\hat{z}$ component, which would mean the ball is moving underground. Thus, Option (B) is incorrect.

### Phase 4: The Final Calculations

After the bounce, the ball acts as a projectile again. Its maximum height $h_1$ depends entirely on its vertical launch velocity $v_z$:

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

Finally, we need to find the horizontal distance $d$ covered during the initial free fall. The time of flight $t$ for a drop of $3h$ is:

$$ t = \sqrt{\frac{2(3h)}{g}} = \sqrt{\frac{6h}{g}} $$

The horizontal distance is simply the constant horizontal velocity multiplied by this time:

$$ d = u_0 t = \sqrt{2gh} 	imes \sqrt{\frac{6h}{g}} = \sqrt{12h^2} = 2\sqrt{3}h $$

The question asks for the ratio $d/h_1$:

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

This confirms that Option (D) is correct.

By carefully breaking the problem into logical phases, we've successfully navigated through energy conservation, projectile kinematics, and collision dynamics to arrive at the correct options: (A), (C), and (D).

Select Answer:

Visualized Solution

$A na l y z in g t h e S e t u p$

$V e l oc i t y a t t h e E x i t o f t h e S l i d e$

$C a l c u l a t in g I ni t ia l V e l oc i t y u_{0}$

$P r o j ec t i l e M o t i o n : T h e F a l l$

$V e l oc i t y J u s t B e f or e I m p a c t$

$A n g l e o f I m p a c t θ$

$T h e B o u n ce : C oe f f i c i e n t o f R es t i t u t i o n$

$V e l oc i t y A f t er t h e B o u n ce v$

$M a x im u m H e i g h t A f t er B o u n ce h_{1}$

$H or i z o n t a l D i s t an ce d$

$R a t i o d / h_{1}$

$C o n c l u s i o n$

The Sigma Insight: Projectile Motion

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$ . As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of $(h / H)$ , the body will take maximum time to reach the ground?

A small ball is thrown from foot of a wall with the minimum possible velocity to hit a bulb B on the ground a distance $L$ away from the wall. Find expression for height $h$ of shadow of the ball on the wall as a function of time $t$ . Acceleration due to gravity is $g$ .

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity at the Exit of the Slide

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Initial Velocity u0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Projectile Motion: The Fall

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity Just Before Impact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle of Impact θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Bounce: Coefficient of Restitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity After the Bounce v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Height After Bounce h1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Distance d

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio d/h1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion