The direction of projection of particle is shown in the figure for an observer on trolley An observer on the ground sees the ball rise vertically The maximum height reached by ball as seen from the trolley is:

10 m

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15 m

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20 m

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5 m

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Solution

The correct option is B 15 m

$10 - v cos 60^{\circ} = 0$

$v = 20 m / s$

$H = \frac{v^{2} {sin}^{2} 60^{\circ}}{2 g} = 15 m$

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Similar questions

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The direction of projection of particle is shown in the figure for an observer on trolley An observer on the ground sees the ball rise vertically The maximum height reached by ball as seen from the trolley is:

The correct option is B 15 m 10−vcos60∘=0 v=20m/s H=v2sin260∘2g=15m

The correct option is B 15 m

$10 - v cos 60^{\circ} = 0$

$v = 20 m / s$

$H = \frac{v^{2} {sin}^{2} 60^{\circ}}{2 g} = 15 m$