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Q.

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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a

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

b

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

c

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

d

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

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 $V_{T} = 3 .5 m / s$

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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 vT and vB respectively, then take ,g=10m/s2