Basic Physics Problem

Discussion in 'After Hours Lounge (Off Topic)' started by Neil J, Sep 9, 2005.

  1. Neil J

    Neil J Stunt Coordinator

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    I'm hoping someone knows how to setup this problem. It's from my physics book but I have a homework problem that is similar. Here goes.
    A rock is dropped off of a cliff and the sound of it hitting below is heard 3.2 seconds later. The speed of sound is 340 m/s. How high is the cliff?
    The answer in the book is 46m.
    Any help would be appreciated.
     
  2. Haggai

    Haggai Producer

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    Let's say that the height of the cliff is y meters, so you're trying to solve for y.

    You drop the rock of the cliff, and it hits the ground some time later, but you don't know how long it takes. Call this time t. You should be able to write down an equation relating t and y.

    You also know that the sound travels back up the cliff--a distance of y meters--at the speed of sound given in the problem. But you don't know how long this takes either. Call this time x. You should be able to get an equation relating x and y.

    Now you have two equations with three variables: x, y, and t. You need another equation relating a couple of those variables to each other. And there's one bit of information given in the problem that hasn't been used yet...the rest I leave to you!
     
  3. Eric Kahn

    Eric Kahn Guest

    you also have to take into consideration the acceleration of the falling object, 16 feet per second, per second as they always used to write it

    you need to calculate on a graph, the acceleration curve of the falling rock and see where it intersects the previously graphed straightline of the constant, which is the speed of sound which is given too you

    I know what you need to do, just don't ask me for equations, I do not know how to make them
     
  4. Brian Perry

    Brian Perry Cinematographer

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    nevermind... [​IMG]
     
  5. BrianW

    BrianW Cinematographer

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    3.2 seconds is the time it took for the rock to fall, plus the time it took for the sound to come back up. The height of the cliff and the fraction of the total time each (rock vs. sound) required are the only unknowns. You can do this with only one equation. The height of the cliff will be a function of the time it takes to hear the rock hit the ground.
     
  6. andrew markworthy

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    You know how they say that you weren't really in the seventies if you can remember anything about it? They're absolutely right. This sort of calculation was routine stuff when I was doing math at school in the seventies. Now I can't begin to work it out. [​IMG]
     
  7. PaulT

    PaulT Supporting Actor

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    The acceleration of a falling obeject is 32ft/sec2 (9.8m/sec2)

    A laden African swallow is another thing [​IMG]
     
  8. Joseph DeMartino

    Joseph DeMartino Lead Actor

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    You would have to if you were calculating this in the real world for some actual engineering reason, but given that this is a exercise in a physics book, and that the rate of acceleration is not one of the parameters given the students, I think this can and should be ignored. If they were meant to take that variable into account they would have been told to and supplied a value for it.

    Regards,

    Joe
     
  9. Haggai

    Haggai Producer

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    But you can't get a numerical value for the answer without having a value for the acceleration due to gravity. Before I posted my hints above, I worked it out using the standard gravity acceleration value that PaulT mentioned in his post, and the answer came out to be the same as in Neil's book.

    I remember that when I took physics in college, the standard value for the acceleration of gravity was given early on in whichever mechanics textbook we were using, and it was then assumed to be the same for every relevant problem after that, unless otherwise specified.
     
  10. RobertR

    RobertR Lead Actor

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    Yes, it's reasonable to assume standard gravitational acceleration, and ignore air drag. I got the same answer as well.
     
  11. Adam Bluhm

    Adam Bluhm Supporting Actor

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    Don't feel bad, Andrew. In 11th grade I ripped through this. I'm now 23 years young and I can't even remember the routine forumulas.

    y=v/t That might be a forumula! (not sure [​IMG] )
     
  12. Tim Hoover

    Tim Hoover Screenwriter

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    Adam - your formula is close! Actually y=vt, where v is the average velocity [​IMG]
     
  13. BrianW

    BrianW Cinematographer

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    I disagree and echo Haggai's comment above. Supplying everything a student needs to solve a problem does little to improve a student's problem-solving skills. The more minimal the given assumptions, the more the student relies on his own thinking ability to fill in the blanks and arrive at an equation that yields the correct answer. Telling students what variables to consider (and, contrapositively, what variables NOT to consider by their omission) will not create new wrinkles in their brains, which is presumably the goal.

    Besides, it's not a variable; it's a constant. [​IMG]
     
  14. Neil J

    Neil J Stunt Coordinator

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    Ok, I'm still working on this problem and several others. Thanks for the help. Also the 9.80 meter/seconds(squared) is given in the book.
     
  15. Adam Bluhm

    Adam Bluhm Supporting Actor

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    Make that a negative 9.8m/s*2! [​IMG]
     

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