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Calculate falling distance when object is already in motion

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data

    A person looking out of a window of a tall building sees a bucket fly past the window at 30 m/s, then hears the bucket hit the ground 8 seconds later. At what altitude is the observer located? Assume negligible air resistance and speed of sound at 340 m/s.

    The problem I am having is that the sound takes an unknown time to arrive at the observer after hitting the ground.

    Using the common kinematic equations and substitution, I have a couple of formulas in the link.

    What are the steps using my equation, if correct? I haven't managed to do it.

    Or, how do I solve it after setting equation#1 equal to velocity * (total time - time for sound to reach observer)?

    I have found various equations for solving similar penny/stone in the well problems , but they all assume a starting velocity of zero.

    As an additional question, what area of algebra should I practice here, in order to solve equations like this?

    2. Relevant equations

    http://s31.postimg.org/d7moph8wr/well.jpg
    upload_2016-4-20_8-12-45.png [Image inserted by moderator]

    3. The attempt at a solution

    The problem I am having is that the sound is already in motion when the time begins, and the sound takes an unknown time to arrive at the observer after hitting the ground due to the unknown distance.
     
    Last edited by a moderator: Apr 20, 2016
  2. jcsd
  3. Apr 20, 2016 #2

    gneill

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    Staff: Mentor

    Hi califauna, welcome to Physics Forums!

    Why don't you write separate equations of motion for the bucket and the sound to begin with? Use different variables for time for each so that both may start from zero. What do you know about the sum of the two times?
     
  4. Apr 21, 2016 #3
    Hi, Im not exactly sure what you mean for the motion of the bucket. I think equation number one is an equation for the motion of the bucket until it it hits the ground isnt it (assuming there is no extra time compenent for the sound to arrive at the listeners ears) ? Regarding the time for sound to reach the observer, would it be something like this :

    Tob(time to reach observer after hiting ground)=altitude/340

    ??
     
    Last edited: Apr 21, 2016
  5. Apr 21, 2016 #4
    by the way, as you may have noticed, the equation I wrote is wrong to start with. I think I should have substituted t (total time to hear sound) for (8-d/340) , not (8-t/340).
     
  6. Apr 21, 2016 #5

    gneill

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    Yes, that's the right idea. Note that the bucket falling and the sound rising both cover the same path (in different directions, of course) so they must cover the same distance equal to the altitude. Call it d.

    To begin with you can treat both motions separately in order to establish their equations of motions.

    Write out the equations for the each. So for example, for the sound the distance d is given by: d = vs*ts, where vs and ts are the speed of sound and the time that the sound takes in moving from the impact site to the observer's ear.

    Do the same for the falling bucket, using a different time variable to represent the time taken for the bucket to pass the observer and finally reach the impact site.

    Finally, you have a third equation that tells you what the sum of those two times must be.
     
  7. Apr 22, 2016 #6
    What is the answer ?
     
  8. Apr 22, 2016 #7

    haruspex

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    Give Califauna time to respond. May be in a different timezone.
     
  9. Apr 24, 2016 #8
    Like this?

    upload_2016-4-24_22-15-53.png
     
  10. Apr 24, 2016 #9

    gneill

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    That's the idea. Make sure you properly identify which "t" is which. There's no subscript on the t in the squared term you wrote.

    What other equation do you have that relates t1 and t2?
     
  11. Apr 24, 2016 #10
    I changed it
     
  12. Apr 24, 2016 #11
    upload_2016-4-25_1-38-38.png
     
  13. Apr 24, 2016 #12

    gneill

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    Okay. Proceed. Solve for the two times.
     
  14. Apr 24, 2016 #13
    I cant . I cant get t1 on its own. I can only get this:
    upload_2016-4-25_3-2-4.png

    Which equation should I start with?
     
  15. Apr 24, 2016 #14

    gneill

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    Start with the one in post #8. Substitute for either t1 or t2 using the relationship in post #11.
     
  16. Apr 24, 2016 #15
    Stuck here:
    upload_2016-4-25_4-40-14.png
     
  17. Apr 24, 2016 #16

    gneill

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    You're doing fine. Hint: Quadratic Formula
     
  18. Apr 27, 2016 #17
    Thanks. Ill get back to this in a few days after going over quadratic formulas again ( and a couple of exams are finished).
     
  19. May 1, 2016 #18
    Solved it using quadratic formula as suggested. Time to reach observer is 1.251 seconds. Total distance from observer is 425.58 meters.

    Thanks again for the help here.
     
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