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A ball is dropped from the top of a building. The ball takes 0.5s 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 at the bottom of the window are V and U respectively then(g=9.8m/s*2)
V+U=12m/s
V-U=9.8m/s
VU=1m/s
V/U=1m/s
Asked by kannanmadavoor3 | 11 May, 2021, 08:11: PM
Expert Answer
If V is speed of ball at top of window and U is the speed of ball at bottom of window
after passing the window in 0.5 s , then we have ,
U = V + (g t )
where g is acceleration due to gravity and t = 0.5 s is time taken to change the speed from V to U .
Hence we have
U = V + (1/2) g ................................(1)
If the window height 3 m is covered by the ball in 0.5 s with initial speed V , then we have
S = V t + (1/2) g t2
where S = 3 m is the window height which is the distance covered by the ball in 0.5 s
Hence we have , 3 = (1/2) V + (1/2) g (1/2)2
3 = ( V / 2 ) + ( g / 8 ) ............................(2)
Let us eliminate g in eqn.(1) and (2) by using substitution g = 2 ( U - V ) from eqn.(1) and
rewrite eqn.(2) as
3 = ( V / 2 ) + [ ( U - V ) / 4 ]
we get, ( V + U ) = 12 m/s from above expression
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