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Question

A body is projected from the top of a tower of height 32 m with a velocity of 20 m/s at an angle of 37∘ above the horizontal. Find the angle below the horizontal with which it hits the ground. (Take g=10 m/s2)

A
tan1(34)
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B
tan1(34)
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C
tan1(74)
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D
tan1(74)
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Solution

The correct option is D tan−1(−74)Horizontal velocity doesn't change. ⇒vx=ux=ucos37∘=16 m/s Initial vertical velocity uy=20sin37∘=12 m/s Displacement at the time of hitting ground sy=−32 m Using third equation of motion, v2y−u2y=2gsy v2y−122=2×−10×−32 ⇒vy=−28 m/s (velocity is downward) If θ is the angle at which the body hits the ground, then tanθ=vyvx=−2816=−74 ∴θ=tan−1(−74) is the angle w.r.t the horizontal at which the body hits the ground.

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