A cave explorer drops a pebble into a cavern to find how far down it is to the water. The stone is heard hitting the water 1.8 s after release. If the speed of sound in the air in the cave is 345 m/s, how deep is the cavern?

The answer is 15 m. How do you get this number using d=4.9(t)^2 and d=v(t)?

well...

put in 345 to d = vt so d = 345t, this is applicable to the part where the sound is going up.

d = 4.9t^2 is applicable to the rock going down, the ts are different.

so now we can add them:

2d = 4.9t_1^2 + 345t_2 \\

and\\

t_1 + t_2 = 1.8

is that enough to help you solve it?