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Velocity Physics Problem

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Two students are on a balcony 23.4 m above the street. One student throws a ball, b1, vertically downward at 15.5 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

    (a) What is the difference in time the balls spend in the air?

    (b) What is the velocity of each ball as it strikes the ground?
    velocity for b1
    velocity for b2

    (c) How far apart are the balls 0.480 s after they are thrown?


    2. Relevant equations

    s = ut + 1/2at
    Not to sure


    3. The attempt at a solution

    I am considering g = 9.8m/s / sec^2.

    Motion of the first ball:

    =>23.4 = (15.5)t + 10(t^2)/2
    =>23.4 = 15.5 t + 5(t^2)
    =>(t^2) + 3.1t - 4.68 = 0
    => t = (-3.1 + 5.3) / 2 = 0.6 sec

    Motion of the second ball

    =>23.4 = - (15.5)t + 10(t^2)/2
    =>23.4 = -(15.5 t) + 5(t^2)
    =>(t^2) - 3.1t - 4.68 = 0
    => t = (3.1 + 5.3) / 2 = 4.2 sec

    Difference in time = 3.6 sec
     
    Last edited by a moderator: Aug 30, 2009
  2. jcsd
  3. Aug 30, 2009 #2

    kuruman

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    Re: Velocity Physics Problem, First one to answer gets 20 dollars!!!

    If you need help with your homework, ask. If have 20 dollars to spend (I hope they are not Zimbabwe dollars), make a donation to PF.
     
  4. Aug 30, 2009 #3

    Doc Al

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    The equation you want is:
    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex]
    Note the the acceleration here is -g = - 9.8 m/s^2.

    What's the initial position, x0? The final position?
     
  5. Aug 30, 2009 #4
    (c) How far apart are the balls 0.480 s after they are thrown?

    I got the first answer
     
  6. Aug 30, 2009 #5

    Doc Al

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    Use the same equation given above.
     
  7. Aug 30, 2009 #6
    Can u fill it in for me at least? Please
     
  8. Aug 30, 2009 #7

    ideasrule

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    One thing to remember is that the two parts of a free-fall trajectory (going up and going down) are symmetrical. If something goes up at 23 m/s, it'll come down at the same speed; if it takes 4 s to reach its maximum point, it takes 4 s to fall down from its maximum point; if it's thrown at 30 degrees to the horizontal, it'll come down at 30 degrees to the horizontal. This is assuming that the ball lands at the same height it was thrown. Knowing this, part (b) is half-answered: the velocity of b2 is 15.5 m/s down, the same speed it was thrown at.

    Now, for part a, this is correct:

    =>23.4 = (15.5)t + 10(t^2)/2
    =>23.4 = 15.5 t + 5(t^2)
    =>(t^2) + 3.1t - 4.68 = 0
    => t = (-3.1 + 5.3) / 2 = 0.6 sec

    but the second part of your solution isn't because displacement is zero; the ball ends up where it starts. You'll get the right answer if you use d=0, but an easier way is to use only the ball's speed and g. Do you know how?

    For part b: think about what acceleration due to gravity means.
    Part c: don't you know how to calculate the position of the balls after a certain amount of time has passed? You've already used the right formula two times.
     
  9. Aug 30, 2009 #8
    (b) What is the velocity of each ball as it strikes the ground?
    I dont know whats the velocity for b1 and b2
     
  10. Aug 30, 2009 #9

    Doc Al

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    You fill it in. What don't you understand? You have x0, v0, a, and t for each throw. Plug in and solve for the position x.
     
  11. Aug 30, 2009 #10

    What does acceleration due to gravity mean? the same thing?
    part c, i dont
     
  12. Aug 30, 2009 #11
    Where do 0.480 s go?
     
  13. Aug 30, 2009 #12

    Doc Al

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    That's the time.
     
  14. Aug 30, 2009 #13
    Im freaking out guys, i cant afford it to get it wrong, 1 more wrong answer and im not able to post any more answers,

    Can u atleast fill it in for b and c and ill do the work
     
  15. Aug 30, 2009 #14

    ideasrule

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    I've pretty much given you all the formulas already. Acceleration due to gravity is what "g" represents; it's the amount that gravity speeds up a falling object per second. It's 9.8 m/s^2, so in one second, 9.8 m/s is added, in 2 seconds, 9.8*2 is added, and in t seconds, gt is added. Consider an object's initial speed and you get the equation v=vo+gt.

    So you know how long the balls spent in the air (from part a), and you know the balls' initial speed. You can solve for v in v=vo+gt.

    For part c, the formula I'm talking about is s = ut + 1/2at. "s" represents displacement.
     
  16. Aug 30, 2009 #15
    Give me the answers for

    (b) What is the velocity of each ball as it strikes the ground?
    velocity for b1 m/s
    velocity for b2 m/s

    (c) How far apart are the balls 0.480 s after they are thrown?


    And ill work my way the hard way until i get the answer
     
  17. Aug 30, 2009 #16
    is 2.832 the correct answer for c?

    Can u give me and fill in the formula for b
     
  18. Aug 30, 2009 #17

    ideasrule

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    Nope. Also, I already gave you the formula for b.
     
  19. Aug 30, 2009 #18
    Can i see the answer for both b and c please, my time is almost over, i have a certain amount of time i need to be finished with
     
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