A stone is dropped from rest from a bridge that is 45 m above a river. A second stone is thrown down 1.o s after the first stone is dropped. The two stones strike the water simultaneously. What was the initial velocity of the second stone?

Answer 1

The first stone:
45 = 4.9t^2, t = 3.03 s. = Fall time.

The 2nd stone:
h = Vo*t + 0.5g*t^2.
45 = Vo*(3.03-1) + 4.9(3.03-1)^2,
45 = Vo*2.03 + 4.9*2.03^2. Vo = ?.