A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3 h 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 \vecu_0=u_0 \hatx 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 \vecv and reaches a maximum height h_1 . The acceleration due to gravity is g and the coefficient of restitution of the ground is 1 / \sqrt3 . Which of the following statement(s) is(are) correct?

\overrightarrowu_0=\sqrt2 g h \hatx, \theta=60^\circ, d / h_1=2 \sqrt3

\overrightarrowu_0=\sqrt2 g h \hatx

Impulse and Momentum PYQ — JEE Advanced 2023 · Shift 1 · Q35 with Solution

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Correct answer: A, C, D

Speed of the ball as it leaves the slide

Since the slide is frictionless, mechanical energy is conserved while the ball moves on it.

The ball starts from rest at height $h$ above the terrace and leaves the slide horizontally at terrace level.

So, $mgh=\frac12 m u_0^2$ Hence, $u_0=\sqrt{2gh}$ Therefore, $\vec u_0=\sqrt{2gh}\,\hat x$

So Option A is correct.

Projectile motion from terrace to ground

The terrace is at height $3h$ above the ground. The ball leaves horizontally, so initial vertical velocity is zero.

Take downward vertical drop from terrace to ground = $3h$ .

For vertical motion: $3h=\frac12 g t^2$ $t=\sqrt{\frac{6h}{g}}$

Horizontal distance travelled: $d=u_0 t=\sqrt{2gh}\cdot \sqrt{\frac{6h}{g}}=2\sqrt{3}\,h$

Just before striking the ground:

horizontal velocity remains $u_x=\sqrt{2gh}$
vertical velocity $u_z=-gt=-g\sqrt{\frac{6h}{g}}=-\sqrt{6gh}$

So the velocity just before collision is $\vec u=\sqrt{2gh}\,\hat x-\sqrt{6gh}\,\hat z$

Angle made with the horizontal at impact

$\tan\theta=\frac{|u_z|}{u_x}=\frac{\sqrt{6gh}}{\sqrt{2gh}}=\sqrt{3}$ Thus, $\theta=60^\circ$

So Option C is correct.

Velocity after bouncing from the ground

For collision with horizontal ground:

horizontal component remains unchanged (smooth ground),
vertical component reverses direction and becomes multiplied by coefficient of restitution $e=\frac1{\sqrt3}$ .

Before collision, vertical speed magnitude is $\sqrt{6gh}$ downward.

After collision, $v_z=e\sqrt{6gh}=\frac1{\sqrt3}\sqrt{6gh}=\sqrt{2gh}$ upward.

Horizontal component after collision: $v_x=\sqrt{2gh}$

Hence,

=\sqrt{2gh}(\hat x+\hat z)$$ So **Option B is incorrect** because it gives a downward vertical component. --- 5. **Maximum height reached after bounce** After bounce, vertical component is $\sqrt{2gh}$ upward. Thus the maximum height above the ground is $$h_1=\frac{v_z^2}{2g}=\frac{2gh}{2g}=h$$ Now, $$d=2\sqrt3\,h$$ So, $$\frac{d}{h_1}=\frac{2\sqrt3\,h}{h}=2\sqrt3$$ So **Option D is correct**. --- 6. **Final conclusion** Correct options are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

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Correct answer: A, C, D

Speed of the ball as it leaves the slide

Since the slide is frictionless, mechanical energy is conserved while the ball moves on it.

The ball starts from rest at height $h$ above the terrace and leaves the slide horizontally at terrace level.

So, $mgh=\frac12 m u_0^2$ Hence, $u_0=\sqrt{2gh}$ Therefore, $\vec u_0=\sqrt{2gh}\,\hat x$

So Option A is correct.

Projectile motion from terrace to ground

The terrace is at height $3h$ above the ground. The ball leaves horizontally, so initial vertical velocity is zero.

Take downward vertical drop from terrace to ground = $3h$ .

For vertical motion: $3h=\frac12 g t^2$ $t=\sqrt{\frac{6h}{g}}$

Horizontal distance travelled: $d=u_0 t=\sqrt{2gh}\cdot \sqrt{\frac{6h}{g}}=2\sqrt{3}\,h$

Just before striking the ground:

horizontal velocity remains $u_x=\sqrt{2gh}$
vertical velocity $u_z=-gt=-g\sqrt{\frac{6h}{g}}=-\sqrt{6gh}$

So the velocity just before collision is $\vec u=\sqrt{2gh}\,\hat x-\sqrt{6gh}\,\hat z$

Angle made with the horizontal at impact

$\tan\theta=\frac{|u_z|}{u_x}=\frac{\sqrt{6gh}}{\sqrt{2gh}}=\sqrt{3}$ Thus, $\theta=60^\circ$

So Option C is correct.

Velocity after bouncing from the ground

For collision with horizontal ground:

horizontal component remains unchanged (smooth ground),
vertical component reverses direction and becomes multiplied by coefficient of restitution $e=\frac1{\sqrt3}$ .

Before collision, vertical speed magnitude is $\sqrt{6gh}$ downward.

After collision, $v_z=e\sqrt{6gh}=\frac1{\sqrt3}\sqrt{6gh}=\sqrt{2gh}$ upward.

Horizontal component after collision: $v_x=\sqrt{2gh}$

Hence,

=\sqrt{2gh}(\hat x+\hat z)$$ So **Option B is incorrect** because it gives a downward vertical component. --- 5. **Maximum height reached after bounce** After bounce, vertical component is $\sqrt{2gh}$ upward. Thus the maximum height above the ground is $$h_1=\frac{v_z^2}{2g}=\frac{2gh}{2g}=h$$ Now, $$d=2\sqrt3\,h$$ So, $$\frac{d}{h_1}=\frac{2\sqrt3\,h}{h}=2\sqrt3$$ So **Option D is correct**. --- 6. **Final conclusion** Correct options are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

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