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Pure Rolling

Maximum heigh...

Question

Maximum height reached by ball $H$ above ground is

\frac{9 h}{35}

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\frac{18 h}{35}

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\frac{18 h}{25}

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\frac{27 h}{125}

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Solution

The correct option is D $\frac{27 h}{125}$
$H = \frac{v^{2} {sin}^{2} sin 37^{\circ}}{2 g}$
$= \frac{(6 g h / 5) (3 / 5)^{2}}{2 g} = \frac{27 h}{125}$

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