A stone is dropped into a well of $20 m$ deep. Another stone is thrown downward with velocity $v$ one second later. If both stones reach the water surface in the well simultaneously $v$ is equal to $(g = 10 m s^{- 2})$

30 m s^{- 1}

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15 m s^{- 1}

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20 m s^{- 1}

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10 m s^{- 1}

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Solution

The correct option is B $15 m s^{- 1}$
First stone was per forming a free fall
so time taken by it to coves the
depth $h = 20 m$ will be
$t = \sqrt{\frac{2 h}{g}}$ just by using $h = \frac{1}{2} g t^{2}$
$t = \sqrt{\frac{2 \times 20}{10}} = 2$ second
Second stone thrown 1 second later
reaches simultaneously with first stone
so time taken will be $t - 1 = 2 - 1 = 1$ second
the second stone will follow
$h = v t_{1} + \frac{1}{2} g t_{1}^{2}$
putting $h = 20 m, 9 = 10 m / s^{2} & t_{1} = 1 s e c$
we get
$20 = v + 5 \times 1 \Rightarrow v = 15 m l s$

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A stone is dropped into a well of 20 m deep. Another stone is thrown downward with velocity v one second later. If both stones reach the water surface in the well simultaneously v is equal to (g=10 ms−2)

A stone is dropped into a well of $20 m$ deep. Another stone is thrown downward with velocity $v$ one second later. If both stones reach the water surface in the well simultaneously $v$ is equal to $(g = 10 m s^{- 2})$