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

The Correct Option is A, C, D

\(u_0=\sqrt{2gh}\)
\(v_z = \sqrt{2g(3h)}\)
\(tan\theta=\frac{v_z}{u}=\sqrt3\)
\(\theta=60^{\circ}\)
\(d=u_0T=u_0\sqrt{2(\frac{3h}{g})}=\sqrt{2gh}\sqrt{2(\frac{3h}{g})}\)
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, \(u_0=\sqrt{2gh},\theta=60^{\circ},\frac{d}{h}=2\sqrt{3}\)