A stone is thrown upwards from a tower with a velocity $50 m s^{- 1}$ . Another stone is simultaneously thrown downwards from the same location with a velocity $50 m s^{- 1}$ . When the first stone is at the highest point, the relative velocity of the second stone with respect to the first stone is (assume that second stone has not yet, reached the ground

Zero

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50 m s^{- 1}

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100 m s^{- 1}

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150 m s^{- 1}

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Solution

The correct option is D $100 m s^{- 1}$
Initially one stone is thrown up with a velocity of $50 m s^{- 1}$
Time it takes to reach stationary.
v =0, u =50, a = -10,
v = u + at
Solving we get 5s.
Meanwhile in 5seconds the other rock initially moving down with u $= 50 m s^{- 1}$ has now attained a velocity v.
$a = + 10 m s^{- 2}$
$v = u + a t = 50 + 5 (10) = 100 m s^{- 1}$
Hence one rock has a velocity of 0, and the other has a velocity of $100 m s^{- 1}$
Hence relative velocity is $100 m s^{- 1}$ option C is correct.

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