A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the ratio of the speeds. The stones would have attained when they hit ground at the base of the cliff.

A man standing on the edge of a cliff throws a stone straight up with

1.	A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the ratio of the speeds. The stones would have attained when they hit ground at the base of the cliff.
Answer» As the stone thrown vertically up will come back to the point of projection with same speed, both stones will move downward with same initial velocity, so both will hit the ground with velocity `v^(2) = u^(2) + 2gh`, i.e., `v = sqrt((u^(2) + 2gh))` So the ratio of speeds attained when they hit the ground is `1 : 1`. [However, the stone projected up will take `(2u//g)` time more to react the ground than the stone projected downwards]

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A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the ratio of the speeds. The stones would have attained when they hit ground at the base of the cliff.

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