Physics
posted by Oscar Zambrano on .
a stone is dropped from a 75m building. when a stone has dropped 15m, a second stone was thrown downward with an initial velocity such that the two stones hit the ground at the same time. what was the initial velocity of the second stone.

d = Vo*t + 4.9t^2 = 15,
0 + 4.9t^2 = 15,
t^2 = 3.06,
t = 1.75s. = Time to fall 15m.
Therefore, the 1st stone had a 1.75s
headstart:
d1 = 0 + 4.9t^2 = 75
t^2 = 15.31,
t1 = 3.91s.
t2 = 3.91  1.75 = 2.16s.
d2 = Vo*t + 4.9t^2 = 75,
2.16*Vo + 4.9(2.16)^2 = 75,
2.16Vo + 22.9 = 75,
2.16*Vo = 75  22.9 = 52.1,
Vo = 24.1m/s. = Initial velocity of 2nd stone 
post