How high is the cliff using speed of sound?

[katamoria]

Homework Statement

To find the height of a cliff, you drop a rock from the top and 10s ater, you hear the sound of it hitting the ground at the foot of the cliff. Ignoring air resistance, how high is the cliff if the speed of sound is 330m/s.

Homework Equations

Xf = Xo = VoT + 1/2aT^2
Vf = Vo + aT
Vf^2 = Vo^2 + 2a(deltaX)

V = (deltaX)/(deltaT) maybe?

The Attempt at a Solution

ok so here is what i tried.
i drew it so that
T1 = time for rock to fall
T2 = time for sound to go back
Tt = Total time = 10s
so
T1+T2 = Tt = 10s

then I did
Vs = velocity of sound, so
Vs = deltaX / deltaT
solved for deltaX which = height of the cliff and got

deltaX = Vs(Tt-T1)

then for the rock,
Vf^2 = Vo^2 +2a(deltaX)
knowing that Vo^2 = 0, and a = g, and solving for deltaX
deltaX = Vf^2/2g

since both equations are solving for the same deltaX, I set them equal to each other
Vs(Tt-T1) = Vf^2/2g

the unknowns are T1, and Vf

I feel pretty good up to this point, then I'm not so sure, but next i tried...
to solve for T2 by
Vf = Vo +at
t is for the rock so...
Vf = g(Tt-T2)

then plug that in and get the equation

Vs(Tt-T1) = g(Tt-T2)^2/2g
but there's still two unknowns with T1 and T2, so...
if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?

am i close? where did i mess up?

 

[anathema]

Your attempt at solving this problem is a good start. However, there are a few things that need to be adjusted in order to get the correct solution.

Firstly, in the equation Vf = Vo + at, the "a" should actually be the acceleration due to gravity, not just "g". So it should be Vf = Vo + gt.

Next, in your equation Vs(Tt-T1) = Vf^2/2g, you have used the wrong value for Vs. The speed of sound is 330m/s, but in this equation, it should be the velocity of the rock, which is unknown and should be represented by Vf. So it should be Vf(Tt-T1) = Vf^2/2g.

Also, in your attempt to solve for T2, you have used the wrong formula. The correct formula for the time it takes for the sound to travel back up to the top of the cliff is T2 = deltaX/Vs, not Vf = g(Tt-T2). So the equation should be T2 = deltaX/Vs.

Finally, you are on the right track with setting T1 equal to Tt-T2, but it should be T1 = Tt-T2, not T1 = (Tt-T2)^2.

So, to summarize, the correct equations to use are:

T1 = Tt-T2
T2 = deltaX/Vs
Vf = Vo + gt
Vf(Tt-T1) = Vf^2/2g

If you plug these equations into each other, you should be able to solve for Vf, which will give you the height of the cliff.

I hope this helps. Good luck with your calculations!

 

[tomcue]

Your approach is on the right track, but there are a few errors in your equations and calculations. First, you are correct in setting up T1+T2 = Tt = 10s, but you should also note that T1 = the time it takes for the rock to fall and T2 = the time it takes for the sound to travel back up to the top of the cliff. Therefore, T2 = 10s - T1. This will be important in your final equation.

Next, you correctly set up the equation Vs(Tt-T1) = Vf^2/2g, but you forgot to include the initial velocity of the rock, which is 0. So the correct equation should be Vs(Tt-T1) = (Vf^2 - Vo^2)/2g.

In order to solve for T2, you can use the equation Vf = Vo + at, but you need to be careful with your subscripts. Vf here represents the final velocity of the sound, not the rock. So the equation should be Vf = Vs + gt. And since we know that T2 = 10s - T1, we can substitute that in and solve for T1.

Once you have solved for T1, you can plug it back into the equation T2 = 10s - T1 to find T2. Once you have both T1 and T2, you can then use them to find the height of the cliff using the equation delta X = VfT2.

Overall, your approach is correct, but just be careful with your equations and make sure to include all necessary variables. Also, remember to double check your units to make sure they are consistent throughout the problem. Good luck!