Amountain climber stands at the top of a 50 meter cliff hanging over a calm pool of water. the climber throws two stone vertically 1 second apart and observes that they cause a single splash when they hit the water. the first stone has a initial velocity of +2.0 meters per second. what will the velocity of each stone be at the instant bitg stones hit the water?

One of two things is true about this question:  EITHER it can't happen
as you've described it, OR you've left out some vital information.

-- IF the first stone was thrown downward with an initial speed and the
second one was dropped from rest 1 second later, then the second one
can never catch up with the first one, and they can never hit the water together.

-- IF the first stone was thrown downward with an initial speed, AND the
second one was released 1 second later, AND they actually do hit the
water together, THEN the second stone must have been given an initial
downward speed greater than 2 m/s, otherwise it could never catch up
with the first one.

Note:
The masses and weights of the stones are irrelevant and not needed.



An afterthought . . . . .

If the first stone was tossed UP at 2 m/s . . . that could be the meaning of the
prominent plus-sign that you wrote next to the 2 . . . then it rises for (2/9.8) second, then begins to fall, and passes the mountain climber's hand on the way down (4/9.8) second after he tossed it, falling at the same 2.0 m/s downward. 

From there, it still has 50m to go before it hits the water. 
50 = 2 T + 1/2 G T²
4.9 T² + 2 T - 50 = 0
T = 3 seconds

The first stone hits the water 3 seconds after passing the mountain climber's hand on the way down at a downward speed of 2.0 m/s.  In that 3 seconds, it gains (3 x 9.8) = 29.4 m/s of additional speed, hitting the water at (29.4 + 2) = 31.4 m/s .

This is all just a guess, assuming that the 2.0 m/s was an UPWARD launch.
Maybe I'll come back later and calculate the second stone.