A stone dropped is from the top of a building and One second later another stone is thrown down with velocity 20m/s.how far below the top will the second over take the first?

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Solution

Let (t) be the time when one overtakes the other. Let (d) be the distance below the cliff where one overtakes the other.

For stone one: d = 1/ g t^2

For stone two: d = 20 (t - 1) + 1/2 g (t - 1)^2. [The time for stone two is LESS by one second because it was thrown one second later.}

1/2 g t^2 = 20 t - 20 + 1/2 g t^2 - g t + 1/2 g

20 - 1/2 g = t (20 - g)

20 - 5 = 10 t

15 = 10 t → t = 1.5 s

d = 1/2 (10) (1.5)^2 → d = 11.25 m

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