The rebound coefficient between a tennis ball | vedclass

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Question

$r=\frac{v_{2}}{v_{1}}=\sqrt{\frac{0.8 \mathrm{H}}{H}}=\sqrt{0.8}$

$\overrightarrow{\mathrm{v}}_{\mathrm{b}}=150 \hat{\mathrm{n}} \mathrm{km} / \ma

Vedclass · Accepted Answer

$323.6$

Vedclass · Answer

$r=\frac{v_{2}}{v_{1}}=\sqrt{\frac{0.8 \mathrm{H}}{H}}=\sqrt{0.8}$

$\overrightarrow{\mathrm{v}}_{\mathrm{b}}=150 \hat{\mathrm{n}} \mathrm{km} / \mathrm{hr}$

$\overrightarrow{\mathrm{v}}_{\mathrm{r}}=-100 \hat{\mathrm{n}} \mathrm{km} / \mathrm{hr}$

$\overrightarrow{\mathrm{v}}_{\mathrm{br}}=250 \hat{\mathrm{n}} \mathrm{km} / \mathrm{hr}$

$\gamma=\sqrt{0.8}=\left|\frac{\overrightarrow{\mathrm{v}}_{\mathrm{br}}}{\overrightarrow{\mathrm{v}}_{\mathrm{br}}}\right|$

$\overrightarrow{\mathrm{v}}_{\mathrm{b}}^{\prime}=\overrightarrow{\mathrm{v}}_{\mathrm{br}}^{\prime}+\overrightarrow{\mathrm{v}}_{\mathrm{r}}$

$=-223.6 \hat{\mathrm{n}}+(-100) \hat{\mathrm{n}}$

$\mathrm{v}_{\mathrm{b}}^{\prime}=-323.6 \hat{\mathrm{n}}$

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