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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/s
Given, 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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