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elessar_iitkgp (2220)

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Olaaa!! Perrrfect answer. 380  [540 rates]

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The car flies off tangentially from the outer circle. Let its speed be v at the moment it flies off the circular path. Then at that moment its at a height h above the ground and the base of the embarkment is at a distance hcosec along the incline.

When the car flies off the embarkment, it flies as a projectile projected horizontally from a height h. The quantity R is measured from the base of the embarkment and not from the point vertically below the point of projection.

Range of the car =  R2 - h2 cot2
Then,
v (2h/g) = (R2 - h2 cot2 )
v = g(R2 - h2 cot2 )/2h



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