A ball thrown horizontally with velocity v from the top of a tower of height h reaches the ground in t seconds. If another ball of double the mass is thrown horizontally with velocity 3v from the top of another tower of height 4h it reaches the ground in (seconds)

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3t/4

t/2

answer is D.

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Detailed Solution

m_{1} = m, u_{1} = v, h_{1} = h, t_{1} = t

$t_{1} = \sqrt{\frac{{2h}_{1}}{g_{1}}}$

$t = \sqrt{\frac{2h}{g}} \to (1)$

$m_{2} = 2 m, u_{2} = 3 v, h_{2} = 4 h, t_{2} = ?$

$t_{2} = \sqrt{\frac{{2h}_{2}}{g_{1}}}$

$= \sqrt{\frac{2 (4h)}{g}}$

$= 2 \sqrt{\frac{2h}{g}}$

$t_{2} = 2 t_{1} \to (2)$

$from (1) & (2)$

$t_{2} = 2 t$

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