# Homework Help: Easy One Dimensional Kinematics

1. Aug 22, 2009

### pointintime

1. The problem statement, all variables and given/known data

A rcok is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the speeds of sound is 340 s^-1 m how high is the cliff.

2. Relevant equations

X = Xo + Vo t + 2^-1 a t^2

3. The attempt at a solution
Ok 340 average velocity and sense there's no acceleration??? then it is also equal to the Vo right?

X = Vo t

plugged in the numbers and got

1156 m which is wrong

my book tells me the answer is 52 m why???

I think my data might be wrong from how i read the problem

I took Vo to be 340 s^-2 m

t to be 3.4 s

acceleration to be zero

2. Aug 22, 2009

### Staff: Mentor

Think of this in two steps:
(1) The rock falls and hits the ocean.
(2) The sound from the splash travels from the ocean to the top of the cliff.
Each takes time, adding up to the total time (which is given).

3. Aug 22, 2009

### pointintime

ok thanks i should be able to do that

4. Aug 22, 2009

### pointintime

ok when I solved for the time that it took for sound to travel

I got

62.41 s or .1852 s

then when I calculated distance

I got

21219.4 m or 62.968 m
and it's obviously the second one but will I know which one is the extraneous soultion because some times it might no be so obvious...

dang looks like i did it wrong to =[

5. Aug 22, 2009

### Staff: Mentor

Show what you did.

6. Aug 22, 2009

### pointintime

ok well i was suppose to get 52 m

does some one know...

um I got .1852 s for the time it takes for the sound to reach the top
i'm taking it that I went wrong there

For my quadratic equation I got the following...

a = 4.90 m/s^2

b = -306.7 m/s

c = 56.64 m

can anyone give me a hand???

y = ax^2 + bx + c

7. Aug 22, 2009

### pointintime

I established...

X = (9.80 m/s^2 (t1^2))/2

x = 340 m/s (t2)

were t1 is the time it took the rock to drop
t2 is the time it took for the time to reach the top of the cliff

t1 + t2 = 3.4 s

t1 = 3.4s - t2

(9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)
(9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)
((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2
19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2) = (680 m/s) t2
453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2= (680 m/s) t2 - (680 m/s) t2
453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2 = 0
453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2
453.2 m + (39.2 m/s^2)t2^2 - (946.6 m/s)t2 = 0
(39.2 m/s^2)t2^2 -(946.6 m/s)t2 + 453.2 m = 0

t2 = (-b +/- sqrt(b^2 - 4ac))/(2a)

t2 = (946.6 m/s +/- sqrt((-946.6 m/s)^2 - 4(39.2 m/s^2)453.2 m))/(2(39.2 m/s^2))
t2 = (946.6 m/s +/- sqrt((-946.6 m/s)^2 - 4(39.2 m/s^2)453.2 m))/(2(39.2 m/s^2))
t2 = (946.6 m/s +/- 907.9 m/s^2)/(2(39.2 m/s^2))
t2 = (946.6 m/s +/- 907.9 m/s^2)/(78.4 m/s^2)

t2 = 23.65 s
t2 = .4936 s

x = Vo t
x = (340 m/s)23.65 s = 8041 m

x = Vo t
x = (340 m/s).4936 s = 167.8 m

Last edited: Aug 22, 2009
8. Aug 22, 2009

### Staff: Mentor

Recall the two steps I mentioned in my earlier post? You need to express the time for each step in terms of the unknown distance. Then solve for that distance.

Let me know when you're done editing your post.

9. Aug 22, 2009

### pointintime

oh I thought i did

10. Aug 22, 2009

### Staff: Mentor

I only know what you tell me. Are you done editing your post?

11. Aug 22, 2009

### Staff: Mentor

Looks good. Solve for t2. You'll get two solutions, only one of which is the one you want.

12. Aug 22, 2009

### pointintime

13. Aug 22, 2009

### Staff: Mentor

My equation is the same as yours. It has two real solutions for t2, only one of which is physically meaningful for this situation.

14. Aug 22, 2009

### pointintime

ok well I finsihed i put my final answers and they were a lot more than 52 m
don't know were i went wrong either

15. Aug 22, 2009

### Staff: Mentor

Show your solution to that quadratic. You must have made a mistake when solving it. (I got 52 m.)

16. Aug 22, 2009

### Staff: Mentor

So far, so good.
Where did that 2 come from on the right hand side?
This (and what follows) is wrong.

Note: Don't keep going back and editing earlier posts--that makes your thinking very hard to follow.

17. Aug 22, 2009

### pointintime

(9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)
(9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)
I multiplied the whole equation by the inverse of the inverse of the 2 to get rid of it
((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2

when I did that i got this
19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2) = (680 m/s) t2

18. Aug 22, 2009

### Staff: Mentor

But you just multiplied the right hand side by 2:
The factor of 2 (in the denominator) still remains on the left hand side. Correct this error.

19. Aug 22, 2009

### pointintime

Is this correct I'll edit it

(9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)

(9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)

((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (340 m/s)t2)2

multiply the whole thing by 2 gave me this

(19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2)) = (680 m/s)t2

20. Aug 22, 2009

### Staff: Mentor

Still wrong. You have an extra factor of 2 on the RHS.

If you wish to multiply by 2, you must multiply both sides by 2.

21. Aug 22, 2009

### pointintime

this is wrong??

(19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2)) = (680 m/s)t2

22. Aug 22, 2009

### Staff: Mentor

Yes. You multiplied the RHS by 2, the LHS by 8.

If you have (blah)/2 = (bleep), then (blah) = 2*(bleep).

23. Aug 22, 2009

### pointintime

so I just leave the left hand side alone

9.8 x 2 = 19.6 i thought i multiplied by 2 on the LHS

24. Aug 22, 2009

### Staff: Mentor

That's once. But you also multiplied the factors within the parentheses (that's twice). And you multiplied once again to remove the 1/2 (that's three times: 2*2*2 = 8).

25. Aug 22, 2009

### pointintime

So is this correct????

(19.6 m/s^2(11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (680 m/s)t2