Easy One Dimensional Kinematics

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Homework Help Overview

The problem involves a rock dropped from a cliff, with the total time taken for the rock to hit the ocean and the sound to travel back to the top being 3.4 seconds. The speed of sound is given as 340 m/s. Participants are tasked with determining the height of the cliff using kinematic equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss breaking the problem into two parts: the fall of the rock and the travel time of sound. There are attempts to apply kinematic equations, but confusion arises regarding the correct interpretation of time and the setup of equations.

Discussion Status

Several participants are exploring different methods to solve the problem, including quadratic equations and kinematic formulas. There is a recognition of potential errors in calculations and setups, with some participants questioning their assumptions and the validity of their results. Guidance has been offered to clarify the steps needed to express time in terms of distance.

Contextual Notes

Participants express uncertainty about the values used for time and acceleration, and there are mentions of extraneous solutions arising from quadratic equations. The discussion reflects a collaborative effort to understand the problem rather than simply finding a solution.

  • #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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