A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m length 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_T and V_B respectively, then $(g = 9 .8 m / s^{2})$ .

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$V_{B} - V_{T} = 1 {ms}^{- 1}$

$\frac{V_{B}}{V_{T}} = 2$

$V_{T} + V_{B} = 12 {ms}^{- 1}$

$V_{T} - V_{B} = 4 .9 {ms}^{- 1}$

answer is A.

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Detailed Solution

$V_{B} = V_{T} + gt = V_{T} + 9 .8 \times 0 .5$
Or $V_{B} - V_{T} = 4 .9$ ….(i)
Further, $V_{B}^{2} = V_{T}^{2} + 2 gs = V_{T}^{2} + 2 \times 9 .8 \times 3$
Or $V_{B}^{2} - V_{T}^{2} = 58 .8$ ….(ii)
Solving Eqs. (1) and (2) we get,
$V_{B} + V_{T} = 12 m / s$