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Mechanics

A ball is dropped from the top of a building. The ball takes 0.5 seconds to fall past the 3m Height of a window some distance from the top of the building. If the speed of the ball at the top and bottom of the window is given by vT and vB respectively, then (g = 9.8m/(sec*sec))

A) vT+vB = 12 m/s
B) vT-vB = 4.9 m/s
C) vBvT = 1m/s
D) vB/vT = 1m/s

Profile image of VASTAV BISWAS
8 Years agoGrade
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1 Answer

Profile image of HIMANSHU SHEKHAR DAS
ApprovedApproved Tutor Answer8 Years ago
here using v=u+a*t, where the symbols have their usual meanings,
we get:
v=VB, u=VT , a= -9.8m/s2 , t= 0.5 sec,
we get VT – VB = 4.9 m/s
 so option B is correct
then using v2- u2=2*a*s
VB2 – VT2= -2*9.8*3
= -58.8
factorising it we get (VB-VT)(VB+ VT) = -58.8 m/s
putting the vlaue of VT – VB = 4.9 m/s we get
VB + VT = 12 m/s
so option A is also correct.
by adding the 2 equations relating VB and VT we can find the value of the velocities and multiplying it or dividing it does not yield option C or D
so option A and B are correct
thanks