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Question

A bomber plane moves horizontally with a speed of 600 m/s and a bomb released from it, strikes the ground in 10 s. The angle with horizontal at which it strikes the ground will be

A
tan1(1/2)
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B
tan1(1/6)
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C
tan1(4/5)
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D
tan1(3/4)
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Solution

The correct option is B tan1(1/6)
Bomb was released so initial velocity will be zero so the equation for vertical motion will be v=gt
putting t=10sec we get the vertical velocity at the moment of hit as v=10×10=100m/s

At the moment of the release the bomb will acquire the horizontal velocity of the plane which will remain constant
as u=600m/s

angle with horizontal θ=tan1VverticalVhorizontal=tan1vu=tan1100600
or θ=tan116

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