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Question

A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the length of a window, the top of the window being at a distance of 3 m from the top of the building. If the speed of the ball at the top and the bottom of the window are VT and VB respectively, then (g=9.8m/sec2)

A
VT+VB=20ms1
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B
VTVB=4.9ms1
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C
VBVT=1ms1
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D
VBVT=1ms1
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Solution

The correct option is A VT+VB=20ms1
From the equations of motion we know that vt2=u2+2as.
here initial velocity"u"=0 so vt=2×9.81×3=7.67m/sec
also,vb=u+at=0+9.81×0.5=4.905m/sec.
so net velocity at the bottom of window will be
VT+VB=7.67+4.905=12.57m/sec

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