Question

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)

Answer 1

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*(t-1)+(1/2)(-g)(t-1)²
Solve for t

(2)
v(first) = 0+(-g)*t
v(second) = 2.08+(-g)*(t-1)