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(C) The horizontal velocity of the bomb is $u = 600 \, km/h = 600 	imes \frac{5}{18} \, m/s = \frac{500}{3} \, m/s$.The height from which the bomb is

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$3.33 \, km$

Vedclass · Answer

(C) The horizontal velocity of the bomb is $u = 600 \, km/h = 600 	imes \frac{5}{18} \, m/s = \frac{500}{3} \, m/s$.The height from which the bomb is released is $h = 1960 \, m$.The time taken by the bomb to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.Taking $g = 9.8 \, m/s^2$,we have $t = \sqrt{\frac{2 	imes 1960}{9.8}} = \sqrt{\frac{3920}{9.8}} = \sqrt{400} = 20 \, s$.The horizontal distance $AB$ is the horizontal displacement,which is given by $AB = u 	imes t$.$AB = \frac{500}{3} 	imes 20 = \frac{10000}{3} \, m = 3333.33 \, m = 3.33 \, km$.

An aeroplane is flying horizontally with a velocity of $600 \, km/h$ at a height of $1960 \, m$. When it is vertically above a point $A$ on the ground,a bomb is released from it. The bomb strikes the ground at point $B$. The distance $AB$ is

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