Linear Motion: calculating height of cliff given speed of sound and time

rahrahrah1

1. The problem statement, all variables and given/known data
A student standing on top of a cliff drops a rock down below into the water and hears it splash 3 seconds later. The speed of sound is 330m/s, what is the height of the cliff


2. Relevant equations
v= d/t
v2 = v1 + at
d = v2-1/2 at2
d = (v1 + v2)t
v2 squared = v1 squared + 2ad
d = v1t + at2


3. The attempt at a solution
when I first saw this, I thought of echos
v=d/t
330 = d/3
d=990
and then divide by two since echo
d= 990/2
d= 495
however, that is horizontal distance not vertical, so I listed my knowns but am unsure as what to do with the 330 m/s

knowns for rock
v1= 0 (since he dropped the rock)
v2=
a = -9.8 m/s2
t = 3s
d =

knowns for speed of sounds
v= 330 m/s
t = 3s
d= ?

then setting the distance of both of these equal to each other and solving for d?

PeterO

Some of the time [most?] is taken up with the rock falling. A small amount at the end is the sound travelling up at 330 m/s.

For example, if you drop something and it falls for 3 seconds, it falls a little less than 45m

At 330 m/s, the sound would take about 0.15 seconds to come back the 45 m

We thus know the cliff is less than 45 m high.

rahrahrah1

hmm, so if I change my variables in the "knowns for rock"
v1: 0
v2:
a: -9.8m/s
t: t-3
d:

and sub in to solve for d...

d rock = v1t + 0.5at squared
= (0)t + 0.5(-9.8)(t-3) squared
= -4.9tsquared + 29.4t -44.1

dsound =vt
= (330)t

330t = -4.9squared + 29.4t -44.1
= -4.9squared -300.6t - 44.1
= [-b +/- √(b squared-4ac)]2a
= [300.6 +/- √(89496)]-9.8
t= -61.2 or -.15
I get the same 0.15 time however, mine is negative for some reason..

PeterO

I think that first time should be 3-t

That change of sign may fix things