A body dropped freely from a height h onto a horizontal plane, bounces up and down and finally comes to rest. The coefficient of restitution is e, the ratio of distance travelled in two consecutive rebounds

1 : e

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e : 1

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1 : e^{2}

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e^{2} : 1

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Solution

The correct option is C $1 : e^{2}$
$\sqrt{\frac{h_{n} + 1}{h_{n}}} = e$
$\Rightarrow h_{1} = e^{2} h$
$h_{2} = e^{4} h$
so $h_{1} : h_{2} = 1 : e^{2}$

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A body dropped freely from a height h onto a horizontal plane, bounces up and down and finally comes to rest. The coefficient of restitution is e, the ratio of distance travelled in two consecutive rebounds

The correct option is C 1:e2√hn+1hn=e⇒h1=e2hh2=e4hso h1:h2=1:e2

The correct option is C $1 : e^{2}$
$\sqrt{\frac{h_{n} + 1}{h_{n}}} = e$
$\Rightarrow h_{1} = e^{2} h$
$h_{2} = e^{4} h$
so $h_{1} : h_{2} = 1 : e^{2}$