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A bomber plane moves horizontally with a speed of 500 ms-1 and a bomb released from it, strikes the ground in 10s. Angle at which it strikes the ground will be (Take, g = 10 ms-2)

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a

tan-115

b

tan-112

c

tan-11

d

tan-15

answer is A.

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Detailed Solution

Horizontal component of velocity, vx = 500 ms-1 and vertical component of velocity while striking the ground,vy=0+10×10=100 ms-1∴ Angle with which it strikes the ground,θ=tan-1vyvx=tan-1100500=tan-115

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A bomber plane moves horizontally with a speed of 500 ms-1 and a bomb released from it, strikes the ground in 10s. Angle at which it strikes the ground will be (Take, g = 10 ms-2)