1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinematics Physics Question

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A caveman drops a rock from ground-level, into a hole. He counts in seconds until he hears the rock hit the bottom of the hole, and finds that the hole is '16 seconds deep.' Assuming there is no air friction, and the initial velocity of the rock is 0m/s, how deep is the hole? Take into account the speed of sound.

    2. Relevant equations
    h = ½g(T1)²
    h = v(T2)
    T1 + T2 = 16 sec
    3. The attempt at a solution

    Attempt 1:
    h = ½g(T1)²
    h=½(9.8m/s²)(16s)²
    h=4.9m/s²(256s)
    h=1254.4m
    h=v(T2)
    1254.4m=340.29m/s(T2)
    1254.4m/340.29m/s=T2
    T2=3.6862s
    h=v(T2)
    1254.4m=v(3.69s)
    1254.4m/3.6862s=v
    v=340.29m/s
    T1+T2=16 sec
    16s+3.6862s = 19.6862s?

    I am confused because the two values do not add up...and than how do i know which variable actually represents the depth of the hole...
     
    Last edited: Feb 11, 2009
  2. jcsd
  3. Feb 11, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    No. The times add to 16 by the statement of the problem.

    h = ½g(T1)² = v(T2)

    and

    T1 + T2 = 16
     
  4. Feb 11, 2009 #3
    Okay, so...
    h = ½g(T1)² = v(T2)

    Therefore...
    h=½(9.8m/s²)(16s)²=v(T2)
    h=4.9m/s²(256s)
    h=1254.4m=340.29m/s(3.6862s)
    h=1254.4m=1254.4m

    But, I don't understand, because the answer can't be 1254.4m...
     
  5. Feb 11, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    You need to stop putting t1 = 16 into your equation.

    The total time is 16.

    You have 2 unknown time variables.

    T1 + T 2 = 16

    Solve the 2 equations for the 2 variables with that in mind.
     
  6. Feb 11, 2009 #5
    Okay, so if I leave T1 as you've said...than that means I have two unknown variables...so how would that even work :confused:
    h = ½g(T1)²
     
  7. Feb 11, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    Eliminate h first of all, it's the same in both directions isn't it?

    ½g(T1)² = v(T2)

    and

    T1 + T2 = 16

    Solve.
     
  8. Feb 11, 2009 #7
    Okay, so...I put what you said

    ½g(T1)² = v(T2)

    and

    T1 + T2 = 16

    And than I sub in the values...

    ½(9.8m/s²)(T1)²=340.29m/s(T2)
    4.9m/s²(T1)²=340.29m/s(T2)
    4.9m/s²/340.29m/s=(T2)/(T1)²
    0.014399s=(T2)/(T1)²
    T1+T2=16

    Than how would I proceed from here, am I supposed to combine the two equations in some way?
     
  9. Feb 11, 2009 #8

    LowlyPion

    User Avatar
    Homework Helper

    You haven't covered substituting variables in your coursework yet?

    T1 = 16 - T2

    ½g(T1)² = v(T2)

    ½g(16 - T2)² = v(T2)

    It could be a long semester.
     
  10. Feb 11, 2009 #9
    Okay, It's not that I can't substitute, its just that there are two variables, so how am I supposed to solve, I'm really confused because if..


    T1 = 16 - T2

    ...How do I even figure out what one of those variables are in the first place?

    Should I assume that T2 = 3.6862s, because 1254.4m/340.29m/s=T2

    So than, 16-3.6862 = 12.3138, so is this my answer?

    The hole is 12.3138m deep?

    But how can this even be, because if I'm not assuming that T1 is even 16s, that messes up my entire equation...
     
    Last edited: Feb 11, 2009
  11. Feb 11, 2009 #10

    LowlyPion

    User Avatar
    Homework Helper

    Simply evaluate the quadratic:

    ½g(16 - T2)² = v(T2)

    Everything is known but T2

    Quadratic Formula maybe?
     
  12. Feb 11, 2009 #11
    ½g(16 - T2)² = v(T2)
    Okay, I get this far..
    4.9m/s²(16-T2)² =340.29m/s

    So, the quadratic equation means [-b+-sqrt(b²-4ac)]/2a...

    Should I assume that 4.9m/s² = c, 340.29m/s = b, and 16 = a? How do I even work with that 16 because its in the bracket with the T2.
     
  13. Feb 11, 2009 #12

    LowlyPion

    User Avatar
    Homework Helper

    You will need to expand it of course.
     
  14. Feb 11, 2009 #13
    Oh right. Expanding.

    So (16-T2)²
    =16²-T2²?

    So is 16² going to be my 'a' value?

    [-340.29m/s+-sqrt(340.29m/s²-4(16²)(4.9m/s²))]/2(16²)

    Would that be my equation?
     
  15. Feb 11, 2009 #14

    LowlyPion

    User Avatar
    Homework Helper

    No.

    (16 -t)(16 - t) is not 16²-t²
     
  16. Feb 11, 2009 #15
    (16 -t)(16 - t).

    Okay, F.O.I.L rules, so..

    96-16t-16t-t²
    =96-t², because the other values cancel out?
     
  17. Feb 11, 2009 #16
    Oh, wait. I THINK I figured it out, can you confirm?

    t = d/343 + √(2d/g) = 16
    solve for d

    d + 343√(2d/(9.8)) = 5488
    d + 155√d = 5488

    let x² = d
    x² + 155x – 5488 = 0
    quadratic equation:
    to solve ax² + by + c = 0
    x = [–b ±√(b²–4ac)] / 2a
    x = [–155 ±√(24025+21952)] / 2
    x = [–155 ± 214] / 2
    skipping the negative answer..
    x = 30

    √d for x
    √d = 30
    d = 900 meters
     
  18. Feb 11, 2009 #17

    LowlyPion

    User Avatar
    Homework Helper

    No.

    256 - 32t + t²

    That equals 343/4.9 * t = 70t = 256 - 32t + t

    t² - 102t + 256 = 0

    T2 = 2.58
    x = 343 * 2.58

    884 = 1/2*g*t² ===> T1 = 13.43

    T1 + T2 = 2.58 + 13.43 ≈ 16
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics Physics Question
Loading...