A ball is dropped from the top of a building. The ball takes 0.5s to fall the 3m length of a window which is located some distance from the top of the building. If the velocities of the ball at the top and at the bottom of the window are v_T and v_B respectively, then $(take, g = 10 m / s^{2})$

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

$V_{B} - V_{T} = 5 {ms}^{- 1}$

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

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

answer is A, B.

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Detailed Solution

$\begin{array}{l} V = u + at \\ V_{B} = V_{T} + 10 (0 .5) \\ V_{B} - V_{T} = 5 \\ S = (\frac{u + V}{2}) t \end{array}$
$\begin{array}{l} 3 = (\frac{V_{T} + V_{B}}{2}) (0.5) \\ V_{T} + V_{B} = 12 \Rightarrow V_{B} = 8.5 m / s \end{array}$
And $V_{T} = 3 .5 m / s$