Easy One Dimensional Kinematics

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The discussion revolves around solving a kinematics problem involving a rock dropped from a cliff, with the total time until the sound of the splash is heard being 3.4 seconds. Participants clarify that the problem requires breaking the scenario into two parts: the time it takes for the rock to fall and the time for the sound to travel back up. Several users attempt to derive the correct equations and solve for the height of the cliff, with one participant initially calculating an incorrect height of 1156 meters. Ultimately, the correct approach leads to the conclusion that the height of the cliff is 52 meters, emphasizing the importance of correctly accounting for both time intervals in the calculations. The discussion highlights common mistakes in solving quadratic equations and the necessity of careful algebraic manipulation.
  • #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
 
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  • #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)
 
  • #36
ok i'll go at it and edit this post

(9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = (340 m/s) t2
(4.90 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (340 m/s) t2
56.64 m + (4.90 m/s^2)t2^2 - (33.32 m/s)t2 = (340 m/s) t2
56.64 m + (4.90 m/s^2)t2^2 - (33.32 m/s)t2 - (340 m/s) t2 = 0
(4.90 m/s^2)t2^2- (373.3 m/s)t2 + 56.64 m
 
Last edited:
  • #37
Is that correcT?
 
  • #38
That's looks much better. Solve it.
 

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