physics
posted by ai .
. An inquisitive physics student and mountain climber climbs a 50.0m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What was the initial velocity of the second stone? (c) What is the velocity of each stone at the instant the two hit the water?

a. Vo*t + 0.5g*t^2 = d.
2t + 4.9t^2 = 50.
4.9t^2 + 2t  50 = 0.
Use Quadratic Formula and get:
t = 2.99681.
b. Vo*3 + 4.9*3^2 = 50.
Vo*3 + 44.1 = 50.
3Vo = 50  44.1 = 5.9.
Vo = 5.9 / 3 = 1.97 m/s.
c. Vf^2 = Vo^2 + 2g*d.
Vf^2 = 2^2 + 19.6*50 = 984.
Vf = 31.4 m/s. = Final velocity.