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…

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 $3 h$ 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 ${\vec{u}}_{0} = u_{0} \overset{\land}{x}$ 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 $\vec{v}$ and reaches a maximum height $h_{1}$ . The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $\frac{1}{\sqrt{3}}$ . Which of the following statement(s) is(are) correct?

Aicharya · Accepted Answer

Using energy conservation, m g h = 1 2 m u 0 2 &#8658; u 0 = 2 g h u 0 is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion v z 2 = 0 2 + 2 - g - 3 h &#8658; v z = 6 g h Angle &#952; can be written as, tan &#952; = v z u = 3 &#8658; &#952; = 60 o Horizontal distance covered just before the collision will be, d = u 0 T = u 0 2 3 h g = 2 g h 2 3 h g = 2 3 h After collision, only velocity along z -direction changes. Therefore, v 1 = e v z = 2 g h and hence v &#8594; = u 0 i &#8743; + v 1 k &#8743; = 2 g h i &#8743; + k &#8743; = 2 g h x &#8743; + z &#8743; Now, maximum height achieved after collision will be, h 1 = v 1 2 2 g = h . Finally, u 0 = 2 g h , &#160; &#952; = 60 o , &#160; d h 1 = 2 3 .

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

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