posted by alex on .
A boy drops a stone-A, down from a cliff 480 meters high, next he throws a stone-B, 1 second later - at what initial velocity was the stone-B be thrown so as to meet the stone-A on ground at the same time? g=10 m/s2.
Require the following:
(g/2)T^2 = 480
(g/2)(T-1)^2 + V*(T-1) = 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.