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Question

A ball is dropped from the top of a building. The ball takes 0.5sec 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 the bottom is VT and VB, the establish a relationship between vb and vt take (g = 9.8m/s).

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Solution

Here using v=u+at , where the symbols have their usual meanings,
We have:
v=VB,

u=VT ,

a= -9.8 m/s2 ,

t= 0.5 sec,

After putting in equation we get

VT – VB = 4.9 m/s

Now using

v2u2=2as

Or,
VB2 – VT2= 2×9.8×3
= -58.8

Or we can write

(VB-VT)(VB+ VT) = -58.8 m/s
putting the value of VT – VB = 4.9 m/s we get
VB + VT = 12 m/s


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