applied physics
posted by jose .
you are standing on a 100m tall buiding above a street and your friend is standing vertically below the building. When you drop a stone, your friend simultaneously throws the identical stone as yours at a speed of 50m/s in the same straight line. Calculate the time taken for the stones to collide

Let the heights of the two stones above the ground be Y1 and Y2.
Y1 = 100  4.9 t^2 (the dropped stone)
Y2 = 4.9 t^2 + 50 t (the thrownupward stone)
Now set Y1 = Y2 and solve for t.
The t^2 terms drop out and you get an easy linear equation to solve.
Enjoy!
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