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Question

# A bomb is released from an aeroplane which is moving horizontally with a velocity 196 m/s at a height of 980 m from the ground. Find the velocity with which the bomb hits the ground.

A
196 m/s
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B
138.5 m/s
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C
240 m/s
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D
144 m/s
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Solution

## The correct option is C 240 m/sGiven, vertical velocity of aeroplane, uy=0 & vertical acceleration ay=g So, using 2nd eqn of motion: (take downward +ve) sy=uyt+12ayt2 980=0+12×9.8×t2 t2=200 or t=14.14 s Velocity in y - direction when bomb hit the ground vy=uy+ayt =0+9.8×14.14 =138.5 m/s Velocity of bomb in x - direction when it hits the ground, vx=196 m/s So, velocity of bomb when it hits the ground v=√v2x+v2y=√(196)2+(138.5)2 ≈√57600 =240 m/s

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