A stone is dropped from the top of a tall cliff, and is later a second stonc is thrown vertically downward with a velocity of \( 20 \mathrm{~ms}^{-1} \). How far below the top of the cliff will the second stone overtake the first?
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If second is thrown at time t0 after first, distance of first from top at time t (measured from throw) is s1 = (1/2) g (t + t0)^2 and of second is s2 = 20 t + (1/2) g t^2. They meet when s1 = s2. Show more…
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