Physics PLEASE HELP!?

Question

PLEASE EXPLAIN/ SHOW WORK!!!
1. A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.36 s after the stone is dropped. How deep is the hole? (m)

2. A stone is dropped into a river from a bridge 120 ft above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. Both stones strike the water at the same time.
(a) What was the initial speed of the second stone? (ft/s)

3. To stop a car, first you require a certain reaction time to begin braking. Then the car slows down at a constant rate. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 79.0 km/h, and 24.4 m when the initial speed is 46.8 km/h.
(a) What is your reaction time?
(b) What is the magnitude of the deceleration?

PLEASE DO ANY ONE OF THEM!
(YOU DONT HAVE TO DO ALL)

WHAT EQUATIONS SHOULD I USE FOR #2???

Answer 1

For the drop, it hits at t1
y(t)=.5*g*t1^2
for the returning sound
y(t2)=343 *t2
t1+t2=1.36 and y(t1) and y(t2) each is the depth of the hole
t2=1.36-t1
and
.5*g*t1^2=343*(1.36-t1)

g=9.81
9.81*t1^2/2+343*t1-343*1.36=0
t1=1.33 seconds
the hole is 8.7 meters deep

2) solve this as
stone 1
y(t)=120-16*t^2
compute t at y=0
t=sqrt(120)/4
=2.74 seconds
and stone 2
y(t)=120-v0*1.74-16*1.74^2

When y(t)=0 they are at he same location at the same time
0=120-v0*1.74-16*1.74^2
solve for v0
41 ft/sec

j