[BigD]

Hello every body I have this one question that I can't seem to solve. I know the answer is 52m but I can't seem to derive this answer even after trying different method and formula. Can some one help me answer it. Thank so much. It need to be solve through kinematic formula.


A rock is dropped from a seacliff and the sound of it stricking the ocean is heard 3.4 seconds later. If the speed of sound is 340m/s, how high is the cliff?


Can someone help me and can you show the steps so I can be able to solve other problems like this one.Thank You

[kp2171]

I cannot get an answer that matches your answer of 52m, but here's my reasoning to the point of getting nowhere, if your answer is correct.

if you are assuming that the person who heard the rock hit is at the top of the cliff, then it isn't as easy as simply (340m/s)(3.4s) = distance.

the reason is that the 3.4s is the total time... with two different processes. Time down is tied to the rate of acceleration due to gravity. Time up is tied to the speed of sound. Does it make sense that the rock will spend a lot more of the 3.4 s falling than the time for the sound to travel back up? So really, the distance for the sound part is (340m/s)(time total - time down)

which is d = (340m/s)(3.4 - time down) = 1156m - (340m/s)time down

okay... so if you want to solve for d then you need to plug into time down.

you also know that d = v(initial)*t + 1/2at^2

plugging in d = (0)t + 1/2(-9.8m/s^2)t^2

= d = 1/2(-9.8m/s^2)t^2

so now you have d expressed in terms of time down by two equations. Combine.
1/2(-9.8m/s^2)t^2 = 1156m - (340m/s)t


okay... looks like a quadratic if you rearrange

-4.9t^2 + 340t - 1156 = 0

solve for quadratic and I get 3.58m or 65.8m as the two roots.

played around with signs too and I don't get your numbers. Oh well...

[dmatos]

kp, when equating d=vt-vtfall to d=1/2a(tfall)^2, you put an extra minus sign in front of the t^2 term. Also, when you solve the quadratic (and get about 3.24s), that gives you tfall, not the distance. You still have to sub back into d=1156-340*tfall to get the distance.

I approached it from a different tack:

ttotal = tfall + tnoise
t = sqrt(2d/a) + d/v
vt - d = sqrt(2d/a)
d^2 - 2vtd + v^2t^2 = 2d/a
d^2 + (2/a - 2vt)d + v^2t^2 = 0
Solve for d directly.

[kp2171]

kp, when equating d=vt-vtfall to d=1/2a(tfall)^2, you put an extra minus sign in front of the t^2 term. Also, when you solve the quadratic (and get about 3.24s), that gives you tfall, not the distance. You still have to sub back into d=1156-340*tfall to get the distance.

I approached it from a different tack:

ttotal = tfall + tnoise
t = sqrt(2d/a) + d/v
vt - d = sqrt(2d/a)
d^2 - 2vtd + v^2t^2 = 2d/a
d^2 + (2/a - 2vt)d + v^2t^2 = 0
Solve for d directly.

thanks for the tips. I threw in the - and took it out at other times when I was working the problems to see if that made a difference... since sometimes you would use it in some problems.

I don't see how setting up for d as I did would have terms of s. everyplace I set up an equation the s terms on bottom were cancelled by those on top. I just don't see it tonight. Maybe tomorrow it'll jump out.

[colbtech]

Look at http://www.haverford.edu/educ/knight...ccelarator.htm

I'm close but my schoolboy (long time ago) maths is not up to the job.