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Projectile Motion

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A bullet is fired horizontally from a height of 4 meters, misses its target, and travels 98.5 meters before striking the ground. Use the above equations to determine the initial muzzle velocity of the gun.


    2. Relevant equations
    Solve for time
    Horizontal velocity


    3. The attempt at a solution

    Not even sure where to start
     
  2. jcsd
  3. May 20, 2008 #2

    Doc Al

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    Staff: Mentor

    What equations do you have to work with?

    Hint: Consider the vertical motion and solve for the time it takes the bullet to fall 4 meters.
     
  4. May 20, 2008 #3
    I know I need to start there, but I guess I am just lost as to how to solve for time. I have the equation, but I just am lost. I have examples in front of me, but I have no idea how they even got some of the numbers I only have the two I gave you
     
  5. May 20, 2008 #4
    . A rock is thrown horizontally from a 115 meter vertical cliff.
    The rock strikes the ground 92.5 meters from the base of the cliff.

    Determine the speed at which the rock was thrown.

    Since we know the value of y, we can begin the problem by solving for the elapsed time. By manipulating the equation algebraically, we solve for t in terms of y:
     
  6. May 20, 2008 #5
    This was the example given to me, but I see through working through it none of the numbers were used to get 4.84, so how do I start?
     
  7. May 20, 2008 #6

    Doc Al

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    Write out the equation that you have. (You are given all the numbers you need.)
     
  8. May 20, 2008 #7
    time = square root of -2y/g right?
     
  9. May 20, 2008 #8
    Here is what I have the vertical component of the velocity
    v y = -gt

    The distance traveled in the -y direction
    y = -1/2 gt^2
     
  10. May 20, 2008 #9
    So to solve for the elapsed time we solve for t in terms of y so

    t = Square root -2y/g
     
  11. May 20, 2008 #10
    So here is where I get lost

    t = square root of 8 meters/9.81 m/s^2 ?
     
  12. May 20, 2008 #11
    Are you still there, or totally frustrated with me? I am so sorry I have been working on papers for two days now and I am exhausted and I cannot get the equations to even post correctly
     
  13. May 20, 2008 #12
    I am just not too sure how in the example given they have square root 230m/9.81m/s^2

    and that equals 4.84s? How do they get that answer?
     
  14. May 20, 2008 #13

    Doc Al

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    Staff: Mentor

    You're on the right track here. The vertical distance fallen as a function of time is given by:

    [tex]y = (1/2) g t^2[/tex]

    Which you can rearrange to solve for the time:

    [tex]t = \sqrt{\frac{2y}{g}}[/tex]
     
  15. May 20, 2008 #14
    Can you tell me what I am missing?
     
  16. May 20, 2008 #15
    Well thanks, but I feel like I am on the wrong track heading off the track to a painful demise!
     
  17. May 20, 2008 #16

    Doc Al

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    Staff: Mentor

    Just use your calculator! You have the expression--just evaluate it.
     
  18. May 20, 2008 #17
    So with my numbers given of the gun being fired at 4 meters and traveling 98.5 meters am I using the 4 meters?

    t = Square root 2 (4)m/?
     
  19. May 20, 2008 #18
    I have used my calculator and feel a bit retarded after an hour and still cannot get the 4.84...
     
  20. May 20, 2008 #19
    I know you have other people to help. I am sorry I was just hoping to get more clarification. This has me totally lost and so I apologize for taking up your time and still not getting it. Thank you for trying though
     
  21. May 20, 2008 #20

    Doc Al

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    Staff: Mentor

    Do it step by step. First: 230/9.8 = ??? What do you get? Then take the square root of that.

    (PF is running comically slow for me tonight. Some pages are taking 5 minutes to open. :mad:)
     
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