physics
posted by alex .
A boy drops a stoneA, down from a cliff 480 meters high, next he throws a stoneB, 1 second later  at what initial velocity was the stoneB be thrown so as to meet the stoneA on ground at the same time? g=10 m/s2.

Require the following:
(g/2)T^2 = 480
(g/2)(T1)^2 + V*(T1) = 480
You have two equations in two unknowns (V and T) and can solve for V.
The first equation tells you that stone A requires
T = 9.8 seconds to fall to the ground. (if g= 10)
Use that value of T in the second equation to solve for V.
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