A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. $(g = 9.8 m / s^{2})$

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Solution

The correct option is B
$S = u \times t = u \times \sqrt{\frac{2 h}{g}} = 60 \times \frac{5}{18} \times \sqrt{\frac{2 \times 490}{9.8}} = \frac{500}{3}$

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Similar questions

Q. A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of

60 {km/h}^{- 1}

and at a height of

490 m

. At the time of dropping the bomb, how far should the aeroplane be from the enemy post so that the bomb may directly hit the target?

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60 {kmh}^{- 1}

and at a height of

490 m

. At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target?

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and at a height of

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g = 9.8 m / s^{2}

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A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. (g=9.8 m/s2)

The correct option is B S=u×t=u×√2hg=60×518×√2×4909.8=5003

A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. $(g = 9.8 m / s^{2})$

The correct option is B
$S = u \times t = u \times \sqrt{\frac{2 h}{g}} = 60 \times \frac{5}{18} \times \sqrt{\frac{2 \times 490}{9.8}} = \frac{500}{3}$