Easy One Dimensional Kinematics

In summary, a rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. Using the equation X = Xo + Vo t + 2^-1 a t^2, the problem is solved by expressing the time for each step (the rock falling and the sound traveling back up) in terms of the unknown distance and solving for that distance. The correct answer is 52 m.
  • #1
pointintime
166
0

Homework Statement



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.

Homework Equations



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

The Attempt at a Solution


Ok 340 average velocity and sense there's no acceleration? then it is also equal to the Vo right?

So i had this

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
 
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  • #2
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
ok thanks i should be able to do that
 
  • #4
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
Show what you did.
 
  • #6
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
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:
  • #8
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
oh I thought i did
 
  • #10
pointintime said:
oh I thought i did
I only know what you tell me. Are you done editing your post?
 
  • #11
pointintime said:
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)
Looks good. Solve for t2. You'll get two solutions, only one of which is the one you want.
 
  • #12
I'm done got nonreal answer...
 
  • #13
pointintime said:
I'm done got nonreal answer...
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
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
pointintime said:
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
Show your solution to that quadratic. You must have made a mistake when solving it. (I got 52 m.)
 
  • #16
pointintime said:
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)
So far, so good.
((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2
Where did that 2 come from on the right hand side?
19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2) = (680 m/s) t2
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
(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
pointintime said:
(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
But you just multiplied the right hand side by 2:
((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2
The factor of 2 (in the denominator) still remains on the left hand side. Correct this error.
 
  • #19
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
pointintime said:
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
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
this is wrong??

(19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2)) = (680 m/s)t2
 
  • #22
pointintime said:
this is wrong??

(19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2)) = (680 m/s)t2
Yes. You multiplied the RHS by 2, the LHS by 8.

If you have (blah)/2 = (bleep), then (blah) = 2*(bleep).
 
  • #23
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
pointintime said:
9.8 x 2 = 19.6 i thought i multiplied by 2 on the LHS
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
So is this correct?

(19.6 m/s^2(11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (680 m/s)t2
 
  • #26
pointintime said:
So is this correct?

(19.6 m/s^2(11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (680 m/s)t2
Sure, but why did you multiply by 2?
 
  • #27
lol is it right... Sure?

I multiplied by two to get rid of the half

did i multiply correctly?
 
  • #28
pointintime said:
lol is it right... Sure?
Yes, it is correct.
I multiplied by two to get rid of the half
The half is still there. You just doubled the 9.8.
did i multiply correctly?
What you did was correct, but served no purpose.
 
  • #29
so what do you recommend i do I did this

19.6 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2) = (680 m/s)t2
226.6 m + (19.6 m/s^2)t2^2 - (133.3 m/s)t2 = (680 m/s)t2
226.6 m + (19.6 m/s^2)t2^2 - (133.3 m/s)t2 - (680 m/s)t2
(19.6 m/s^2)t2^2 - (813.3 m/s)t2 + 226.6 m
t2 = (-b +/- (b^2 - 4ac)^.5)/(2a)
t2 = (813.3 m/s +/s ((-813.3 m/s)^2 - 4(19.6 m/s^2)226.6 m)^.5)/(2(19.6 m/s^2)

I got

41.21 s and .2805 s

X = Vo t

X = (340 m/s).2805 s

gave me

95.37 m

please tell me were i went wrong
 
  • #30
This is correct:

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

But you solved this:

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

I suggest leaving your units out of the equation. They just clutter things up and make it harder to spot mistakes. Also, the X2 and X2 icons on the toolbar allow you to do superscripts and subscripts. This is much easier to read:

19.6(11.56+t22-6.8t2)/2=680t2
 
  • #31
wait i thought i was suppose to take the two out

this is wrong

19.6 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2) = (680 m/s)t2
 
  • #32
pointintime said:
so what do you recommend i do I did this

19.6 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2) = (680 m/s)t2
This is still not correct. You multiplied the LHS by 4 and the RHS by 2.
 
  • #33
Could you please show me what to do...

sorry =[
 
  • #34
I do not know what to do...
 
  • #35
pointintime said:
Could you please show me what to do...

sorry =[
You have something like this:
9.8*(blah)/2 = (bleep)

To get rid of the 2, multiply both sides by 2:
9.8*(blah) = 2*(bleep) (not 19.6*(blah))

Or just use:
(9.8/2)*(blah) = (bleep)
4.9*(blah) = (bleep)
 

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