Kinematics Question 420

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

Options:

A) vT+vB=12ms1

B) vTvB=4.9ms1

C) vBvT=1ms1

D) vB/vT=1ms1

Show Answer

Answer:

Correct Answer: B

Solution:

vB2=vT2+2(10)(3)
vB2=vT2+60 .(i)

Also, vB=vT+(10)(0.5) .(ii) Solve eqs. (i) and (ii) We get vTvB=4.9 m/s



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