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

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?
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Recognitions:
Homework Help
 Quote by 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?
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.
 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..

Recognitions:
Homework Help

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

 Quote by 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..
I think that first time should be 3-t

That change of sign may fix things

 Tags calculating height, linear motion, speed of sound, time