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A bomb is dropped on an enemy post on the ground by an aeroplane flying horizontally with a velocity of $60 {kmh}^{- 1}$ at a height of 490 m. At the time of dropping the bomb, the horizontal distance of the aeroplane from the enemy post so that the bomb hits the target is

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$(\frac{1700}{3}) m$

$(\frac{400}{3}) m$

$(\frac{500}{3}) m$

$498 m$

answer is B.

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Detailed Solution

$\begin{array}{l} S = u \times t = u \times \sqrt{\frac{2 h}{g}} = 60 \times \frac{5}{18} \times \sqrt{\frac{2 \times 490}{9.8}} \\ = \frac{500}{3} m \end{array}$

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