Please help with physics problem!!!?

Question

A mountain climber stands at the top of a 50.0 m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash. The first stone had an initial velocity of -1.60 m/s.
(a) How long after release of the first stone did the two stones hit the water?
_____s
(b) What initial velocity must the second stone have had, given that they hit the water simultaneously?
_____m/s
(c) What was the velocity of each stone at the instant it hit the water?
first stone ____m/s
second stone _____m/s

Answer 1

a) The answer is t1 in the equation
y = Vo*t1 + (1/2)*g*(t1)^2
where
y = +50 m
Vo = -1.60 m/s
g = -9.8 m/s^2
It's a quadratic, one value for t will be negative - ignore that one.

b) y = Vo*t2 + (1/2)*g*(t2)^2
where
y = +50 m
Vo = the unknown
g = -9.8 m/s^2
t2 = t1-1

c) Vf^2 = Vo^2 + 2*g*y
You know Vo, g, and y