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A ball A is projected from the ground such that its horizontal range is maximum. Another ball B is dropped from a height equal to the maximum range of A. The ratio of the time of flight of A to the time of fall of B is

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$\sqrt{2}$ : 1

1 : 2

1 : 1

1 : $\sqrt{2}$

answer is C.

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Detailed Solution

$θ$ = 45^o as R_max and $R = \frac{u^{2}}{g}$
$T_{A} = \frac{2 usin 45}{g} = \frac{\sqrt{2} u}{g}; T_{B} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 R}{g}}$
as free fall body $= \sqrt{\frac{2 u^{2}}{g^{2}}} = \frac{\sqrt{2} u}{g}$
T_A : T_B = 1 : 1

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