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A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is dropped vertically down 1s after the first is dropped. Both stones strike the water at the same time.

a) What is the initial speed of the second stone?


Write an equation for the vertical height of each stone. Let t=0 be the time the first stone is dropped. The second stone must have an initial vertical (downward) velocity so it can catch up with the first.
Y1 = 43.9 - (g/2) t^2
Y2 = 43.9 - Vo (t-1) - (g/2)(t-1)^2

Set Y1 = 0 and Y2 = 0 (the heights when each hit the ground). Use the first equation to solve for t. Use the second equation to solve for Vo, the initial downward velocity of the second stone. Note that t-1 is the time of flight of the second stone.

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