posted by brizett .
While standing on a bridge 30.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.
d1 = 0.5g*t^2 = 30-3.2 = 26.8.
4.9t^2 = 26.8
t^2 = 5.47
t = 2.34 s. = Fall time of 1st stone.
The 2nd stone must fall 30 meters in
d2 = Vo*t + 0.5g*t^2 = 30 m.
Vo*2.34 + 4.9(2.34)^2 = 30
2.34Vo + 26.83 = 30
2.34Vo = 30-26.83 = 3.17
Vo = 1.35 m/s. = Initial velocity of 2nd stone.