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Physics Q

Bldr_DH

Monkey
Aug 8, 2003
767
0
NO BO CO
I just have question about a physics problem -- I just need to know how to solve it, not the actual answer. Here's the question:

A stone is dropped from a sea cliff and the sound of it hitting the ocean is heard 3.4s after it's release. If the speed of sound is 340 m/s, how high is the cliff.

Basically, I know the distance down, and the distance back up again are the same, and the the acceleration of the rock is -9.8m/s/s (gravity). I've tried setting the distances equal to each other, then plugging in the numbers to get the time it takes the sound to travel the distance, then multiply that number by 340m/s to get the distance, but alas that hasn't worked.

I'm sort of at a loss as to what I should do... :help:
 

CrabJoe StretchPants

Reincarnated Crab Walking Head Spinning Bruce Dick
Nov 30, 2003
11,569
422
Groton, MA
the 340m/s is extra information, and isnt needed. use the displacement/acceleration relationship: d= v(inital)t + (1/2)at^2

solve for d,
v(initial)=0 (it was dropped not thrown)
t=3.4s
a=-9.8m/s/s

you should get 56.644m
 

Bldr_DH

Monkey
Aug 8, 2003
767
0
NO BO CO
Oh, ok, so since it's so fast I should essentially just ignore it for such a small distance... makes sense. That makes things easy... Thanks.
 

Bldr_DH

Monkey
Aug 8, 2003
767
0
NO BO CO
hrm... apparently 56.644m isn't quite right, so I'm assuming they took into account the time it took for the sound to come back up, and removed it from the time it took to get down, thus the distance is smaller. The answer is 52, so you weren't far off, but I fear the teach will scrutinize for accuracy... What do you think? Any other options?
 

bomberboy11

Monkey
Jul 15, 2005
665
0
At a computer...duh
If they want the accuracy to be to 3 decimal places, chances are they want you to be factoring in the speed of sound as well. In that case it becomes a piece-wise function, using variables T_down and T_up (time it takes for the object to go down and hit the bottom and time it takes for the sound to come up back to the viewer) where T_down + T_up = 3.4 seconds. You will have functions d = (1/2)*a*(T_down)^2 = 340*T_up. From here, make the substituion T_up = 3.4 - T_down (you can isolate either time variable, but this makes it simpler because of the square in the acceleration function). Solve (1/2)*a*(T_down)^2 = 340*(3.4 - T_down) for T_down (this will likely have to be done on your calculator using either the graphing method of a binomial where it intersects 0 or just the quadratic formula - in either case, use the positive value), then plug that back into d = (1/2)*a*(T_down)^2 to get your distance. Hope this helps if it turns out that the speed of sound turns out not to be negligible.
 

beestiboy

Monkey
May 21, 2005
321
0
Merded, ca
Wow i forgot a lot of calc after 5 years since leaving college.

so i guess they are discrediting any cross wind or off shore flow. They always simplify it to the point where it is no longer a practicle question.(j/k) The answer is who cares so long as you hear the splash and dont fall off the cliff.
 

Bldr_DH

Monkey
Aug 8, 2003
767
0
NO BO CO
Yep, it checked out... It's what I was doing, but I was substituting it differently -- made it more complicated and I ended up making a mistake with positive/negative signs somewhere. Thanks to those who helped 'n such.