physics
posted by Jamie on .
An inquisitive physics student and mountain climber climbs a 50.0 m 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 2.08 m/s downward.
(a) How long after release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if they are to hit simultaneously?
2 m/s (downward)
(c) What is the speed of each stone at the instant the two hit the water?
first stone? (downward)
the second stone? (downward)

I'd say they would never hit the water at the same time because the first stone had a initial DOWNWARD velocity of 2.08 m/s. The second stone lost one second plus it started at zero vertical velocity.
If the situation is reversed, if the second stone is given an initial downward velocity of 2.08 m/s, there is a chance they may hit the water together.
(1)
Using
S=vertical distance travelled, m
t = time in seconds since release of first stone.
S = 0*t+(1/2)(g)t² = 2.08*(t1)+(1/2)(g)(t1)²
Solve for t
(2)
v(first) = 0+(g)*t
v(second) = 2.08+(g)*(t1)