KinematicsHard

Question

A ball is dropped from the top of a building. The ball takes The ball 0.5 s to fall the 3m length of a window some distance from the top of the building. If the velocities of the ball at the top and the bottom of the window are vT and vB respectively, then (take g = 10 m/s2) : -

Options

A.vT + vB  = 12 ms-1   
B.vB - vT = 4.9 ms-1
C.vBvT = 1 m-1
D. = 1 m-1

Solution


vB = vT + 9.8 × 0.5 = vT + 4.9
vB - vT = 4.9 m/s and
vB2 - vT2 = 2gs = 2 × 9.8 × 3 = 58.8
⇒ (vB + vT) × (vB - vT) = 2 × 9.8 × 3
⇒ vB + vT = 12 m/s

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