A ball falls freely from rest from a height of 6.25 m on to a hard horizontal surface. If the ball reaches a…

A ball falls freely from rest from a height of 6.25 m on to a hard horizontal surface. If the ball reaches a height of 81 cm after second bounce from the surface, the coefficient of restitution is
  1. 0.3
  2. 0.45
  3. 0.75
  4. 0.6

Solution


From the figure, $\begin{aligned} & u=\sqrt{2 \times 10 \times 6.25} \\ & 5 \sqrt{5} \mathrm{~m} / \mathrm{s} \\ & v^{\prime}=\sqrt{2 \times 10 \times 0.81}=\frac{9 \sqrt{5}}{5} \mathrm{~m} / \mathrm{s} \end{aligned}$
Coefficient of restitution, $\begin{aligned} & \mathrm{e}^2=\frac{\mathrm{v}^{\prime}}{\mathrm{u}}=\frac{\frac{9 \sqrt{5}}{5}}{5 \sqrt{5}}=\frac{9}{25} \\ & \therefore \quad \mathrm{e}=0.6 \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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