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 $\hat{u}_0=u_0\hat{x}$ 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 of v and reaches a maximum height of ℎ1. The acceleration due to gravity is 𝑔 and the coefficient of restitution of the ground is $\frac{1}{\sqrt3}$. Which of the following statement(s) is(are) correct?

• A. $\vec{u}_0=\sqrt{2gh}\hat{x}$
• B. $\vec{v}=\sqrt{2gh}(\hat{x}-\hat{z})$
• C. $\theta=60^{\circ}$
• D. $\frac{d}{h_1}=2\sqrt3$

Answer 1

Answer: (A), (C), (D)

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

Answer 2

$u_0=\sqrt{2gh}$
$v_z = \sqrt{2g(3h)}$
$tan\theta=\frac{v_z}{u}=\sqrt3$
$\theta=60^{\circ}$
$d=u_0T=u_0\sqrt{2(\frac{3h}{g})}=\sqrt{2gh}\sqrt{2(\frac{3h}{g})}$
Velocity after the collision,
$v_1=ev_z=\sqrt{2gh}$
$\vec{v}=v_1\hat{k}+u_0\hat{i}$

$=\sqrt{2gh}[\hat{i}+\hat{j}]$

$h_1=\frac{v_1^2}{2g}=h$
Finally, $u_0=\sqrt{2gh},\theta=60^{\circ},\frac{d}{h}=2\sqrt{3}$