Physics HElP!!!!!!!!!!!!!?

Question

A mountain climber stands at the top o a 59.0 meter cliff hanging over a calm pool of water the climber throws two stones vertically 1.7 seconds apart and observes that they cause a single splash when they hit the water the first stone has an initial velocity of +2.3m/s.
what will be the velocity of the first stone be at the instant both stones hit the water? answer in m/s.

how long after the release of the first stone will the tow stones hit the water? answer in units of seconds.

What is the initial velocity of the second stone when it is thrown? answer in units of m/s

what will the velocity of the second stone be the instant both stones hit the water? answer im m/s

Answer 1

the equations are
y1(t)=59+2.3*t-.5*g*t^2
y2(t)=59+v0*(t-1.7)-
.5*g*(t-1.7)^2, t>=1.7

let's look at stone 1 to determine when it hits the water
0=59+2.3*t-.5*g*t^2
using g=9.81, the positive root is
3.7 seconds
(note, the fact that a negative root exists can be explained: Since I chose an arbitrary time interval, there could have been an event where the stone was shot up from the water and reached the point where the climber threw the stone at t=0 with a vertical velocity of 2.3 m/s)

since they both arrive at y(t)=0 at the same instant
for stone 2
0=59+v0*(3.7-1.7)-
.5*g*(3.7-1.7)^2

0=59+v0*2-2*9.81
v0=-19.7 m/s

he must throw the stone downward to catch up to the first

let's check this:
At what time does the first stone fall below the cliff?
y(t)=59
=59+2.3*t-.5*g*t^2
t=0 since that is the moment it was thrown, and
t=4.6/9.81
=.47 seconds

since the second stone was thrown well after the first stone passed below the cliff, it had to be thrown downward to catch up.

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