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Stone ‘A’ is projected vertically upwards with an initial velocity $9.8 m s^{- 1}$ from certain height. At the same time another stone ‘B’ is dropped from the same height. Relative velocity of the stone B with respect to stone ‘A’ after two seconds is

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$9.8 m s^{- 1}$

$19.6 m s^{- 1}$

$4.9 m s^{- 1}$

$29.4 m s^{- 1}$

answer is A.

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Detailed Solution

$t_{A} = \frac{u}{g} = \frac{9.8}{9.8} = 1 s V_{A} = u_{A} - g t = 9.8 - 9.8 \times 2 = - 9.8 m s^{- 1}$

$∴ V_{A} = 9.8 m s^{- 1}$ downwards

$V_{B} = u_{B} + g t = 0 + 9.8 \times 2 = 19.6$ downwards

$V_{B A} = V_{B} - V_{A} = 9.8 m s^{- 1}$

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