A particle is projected from a stationery trolley. After projection, the trolley moves with a velocity $2 \sqrt{15} m / s$ . For an observer on the trolley, the direction of the particle is as shown in figure while for the observer on the ground, the ball rises vertically. The maximum height reached by the ball from the trolley is $h$ meter. The value of $h$ will be_______.

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Solution

$v_{x} = v cos 60^{o} = 2 \sqrt{15} m / s$
$∴ v = 4 \sqrt{15} m / s$
$∴ v_{y} = v sin 60^{o} = 4 \sqrt{15} \frac{\sqrt{3}}{2} = 2 \sqrt{45} m / s$
$H_{m a x} = \frac{v_{y}^{2}}{2 g} = \frac{4 \times 45}{20} = 9 m$

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vx=vcos60o=2√15m/s∴v=4√15m/s∴vy=vsin60o=4√15√32=2√45 m/sHmax=v2y2g=4×4520=9m

$v_{x} = v cos 60^{o} = 2 \sqrt{15} m / s$
$∴ v = 4 \sqrt{15} m / s$
$∴ v_{y} = v sin 60^{o} = 4 \sqrt{15} \frac{\sqrt{3}}{2} = 2 \sqrt{45} m / s$
$H_{m a x} = \frac{v_{y}^{2}}{2 g} = \frac{4 \times 45}{20} = 9 m$