If a ball which is dropped from a height of 2.25 m on a smooth floor attains the height of bounce equal to 1.00 m, the coefficient of the restitution between the ball and the floor is equal to:

Concept:

When the ball is dropped from some height (h) the velocity with which it impacts the surface is given by \(v = \sqrt {2gh} \)

Coefficient of restitution = \(\frac{{Velocity\;of\;Separation}}{{Velocity\;of\;Approach}} = \frac{{\left| {\left( {{v_2} - {v_1}} \right)} \right|}}{{\left| {\left( {{u_1} - \;{u_2}} \right)} \right|}}\)

u1 and u2 are the initial velocities of the ball and the surface respectively before impact

v1 and v2 are the velocities of the ball and the surface respectively after the impact

Calculation:

Before impact, the velocity is given by \({u_1} = \sqrt {2gh}\)

The initial and final velocity of the surface will be Zero

After the first impact \(e = \frac{{\left| {\left( {0 - {v_1}} \right)} \right|}}{{\left| {({u_1} - 0)} \right|}} = \frac{{{v_1}}}{{{u_1}}}\)

\(e = \frac{{\sqrt {2g{h_1}} }}{{\sqrt {2gh} }} = \sqrt {\frac{{{h_1}}}{h}}\)

Calculation:

Given:

h₁ = 1 m, h = 2.25 m

\(e=\sqrt{\frac{1}{2.25}}=0.66\)