A particle is projected upwards with some velocity. At what height from ground should another particle be just dropped at the same time so that both reach the ground simultaneously. Assume that first particle reaches to a maximum height $H$ .

3 H

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4 H

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5 H

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6 H

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Solution

The correct option is B $4 H$
Let the height from which the second particle is dropped be $h$ .
Time of flight of second particle $T = \sqrt{\frac{2 h}{g}}$
Since, both particles hit the ground simultaneously. So both must have equal time of flight.
Time of flight of first particle $T = \frac{2 u sin θ}{g}$
Using equation for maximum height $H = \frac{u^{2} {sin}^{2} θ}{2 g}$
We get $H = \frac{g}{8} \times T^{2} = \frac{g}{8} \times \frac{2 h}{g}$
$⟹ h = 4 H$

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