If the speed of sound is 340m/s, how high is the cliff?

[sundrops]

Physics Help Needed! (work shown just need an extra push in the right direction! )

Hi, I'm currently taking a Physics 100 level course, so this question should be no problem for you smart people out there. lol My grade 12 teacher definitely did not prep me for this semester - or maybe I just don't remember my physics. Either way I'm having lots of trouble and was hoping someone would give me a hand.

Here's the problem: A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2s later. If the speed of sound is 340m/s, how high is the cliff?

Well what I did is I used the following formula:
d = 0.5(Vf + Vi)t
d = 0.5(340m/s - Om/s)(3.2s)
d = 544m

but that seems too simple of a problem to me - not to mention the GINORMOUS height I got for the cliff. It just doesn't seem reasonable! help!

 

[sundrops]

please help!

 

[dextercioby]

You can find the same problem in this forum.Can u solve a quadratic?

It involves knowing the speed of sound in air...

Daniel.

 

[sundrops]

The speed of sound is given. I need to know how to interpret the data given to me - - because obviously I've done something very wrong.

 

[dextercioby]

Oh,yes u did...Consider the fact that the total time (3.2s) is the sum between the time of the free fall and the time it takes for the sound to come back your ear...

HINT:For the 2 times,the distance is the same and is unknown...

Daniel.

 

[sundrops]

I honestly don't know - I know that you're trying to help me but I just don't understand - I'm not given the velocity of the falling rock - nor am i given the time that it takes for the rock to fall and hit the water - it doesn't seem like there is enough information to solve the problem.

the only knowns I have are:

a of rock = -9.8m/s^2
v of sound = 340m/s
total time = 3.2s
d = ?

 

[learningphysics]

I honestly don't know - I know that you're trying to help me but I just don't understand - I'm not given the velocity of the falling rock - nor am i given the time that it takes for the rock to fall and hit the water - it doesn't seem like there is enough information to solve the problem.

the only knowns I have are:

a of rock = -9.8m/s^2
v of sound = 340m/s
total time = 3.2s
d = ?

Let t1 be the time it takes for the rock to drop to the bottom.
Let t2 be the time it takes for the sound to travel back to the top.

Can you come up with some equations using t1, t2 and d?

 

[sundrops]

so something like:
t1 + t2 = 3.2, 3.2-t1=t2 (and vise-versa)

and then use

x = xo + (vo)(t1) + 0.5(a)(t2)^2
x = 0 + 0 + 0.5(-9.8m/s^2)(3.2 - t2)^2
x = -15.68 - (-4.9)t2

and

x = (Vf + Vi)t2
x = (340m/s^2 + 0m/s^2)(3.2 - t1)
x = 1088m/s - (340)t1

like that? and then what?
man this is getting frusterating!

 

[sundrops]

wow 57 views and only 7 posts! hehe please help me! I'm really stuck!

 

[futb0l]

Ooops\, sorry ignore this. :P

 

[sundrops]

i'm really confused

 

[futb0l]

Hmm, I am not sure if this is right, but here's a try:

$$t_1 + t_2 = 3.2$$

...

$$v = v_0 + gt_1$$

rearrange so that

$$t_1 = \frac{v - v_0}{g}$$

using work energy theorem

$$v = \sqrt{2gh}$$

and you also know that

$$t_2 = \frac{h}{v_s}$$

So then you can solve for h.

 

[learningphysics]

so something like:
t1 + t2 = 3.2, 3.2-t1=t2 (and vise-versa)

and then use

x = xo + (vo)(t1) + 0.5(a)(t2)^2

This formula should only have one time: t1.

x= vo*t1 + 0.5*a*t1^2

Since v0=0, this becomes

x= 0.5*(9.8)t1^2. (Since I'm only dealing with distances here and since the rock is dropping in the downward direction I use +9.8 instead of -9.8)

x = (Vf + Vi)t2

I don't know where you got this formula... did you mean
x=0.5(Vf+Vi)t2 ? You can use that. Remember here where are talking about the distance traveled by the sound back upward, and the speed of sound is constant.
so Vf=Vi=340.

Or the easier way is to just use
x= v*t2
x=340*t2

Ok. So there are 3 equations you've got:

t1+t2=3.2
x=0.5*9.8*t1^2
x=340*t2

Can you solve for x using these 3 equations? You've got 3 equations in 3 unknowns.

 

[sundrops]

wow! that should definitely help - I'll get back to you once I've tried again - thanks!

 

[sundrops]

okay using those three equations:

i) t1 + t2 = 3.2s
ii) x=0.5(9.8m/s^2)(t1)^2
iii) x = 340m/s(t2)


ii) x = (4.9m/s^2)(t1)^2

iii) x = 340m/s(3.2s - t1)
x = 1088m - 340m/s(t1)

so now I need to somehow combine the two t1's and solve for x

would that work? or am I way off track?

 

[sundrops]

also how would i combine the t1's ?

 

[sundrops]

it seems really complicated becuz in one equation the t1 is squared

 

[sundrops]

learningphysics are you still around?

 

[sundrops]

I'm sooo close to being done guys - please can someone explain how to combine the two t1 values so that I can solve for x - and thus solving this problem?

 

[learningphysics]

okay using those three equations:

i) t1 + t2 = 3.2s
ii) x=0.5(9.8m/s^2)(t1)^2
iii) x = 340m/s(t2)


ii) x = (4.9m/s^2)(t1)^2

iii) x = 340m/s(3.2s - t1)
x = 1088m - 340m/s(t1)

so now I need to somehow combine the two t1's and solve for x

would that work? or am I way off track?

I'd solve for t1 in equation iii (so get t1 in terms of x) and then plug it into equation ii. So you get one equation all in x.

Yes, it is kind of complicated... but stick with it. Let me know what you get.

EDIT: Much better to solve for t1 in equation iii, and then plug into equation ii. Makes things easier. No square root.

Then solve the quadratic!

 

[sundrops]

oops! I just kept trying and trying and I plugged it into equation iii) before having read your post. anywhos - I then used the quadratic equation to find t1 = 3.06s and then I plugged t1 into equation ii) and found x = 46.02m

Does that sound about right?

 

[sundrops]

lol I just read your edit - awesome! so that's what I did - is my answer correct?

 

[sundrops]

*crosses her fingers in hopes the answer is correct and that she may finally be done her physics homework*

 

[learningphysics]

oops! I just kept trying and trying and I plugged it into equation iii) before having read your post. anywhos - I then used the quadratic equation to find t1 = 3.06s and then I plugged t1 into equation ii) and found x = 46.02m

Does that sound about right?

That sounds about right to me! Did the computer accept your answer?

 

[sundrops]

YES!

thanks soooo much for your help! You've been awesome! :D

 

[learningphysics]

YES!

thanks soooo much for your help! You've been awesome! :D

You're welcome.

 

[futb0l]

Hmm.. My final equation would be:

$$\sqrt{\frac{2h}{g}} + \frac{h}{v_s} = 3.2$$

Edit: Ooops, too late, you already solved it. Sorry, I didnt Refresh.