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Finding Initial Velocity w/o Time to clear a specific height

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A basketball player shoots the ball at a height of 1.7m at an angle of 60 degrees into a hoop. The range is 10ft or 3.05m. The height from the ground to the top of the hoop is 10ft or 3.05m. What is the initial velocity m/s to make it to the hoop?



    2. Relevant equations

    1) Vy^2 = Viy^2 - 2g(Change in Y)
    2) Y = Yi + Viy(t) - 1/2gt^2
    3) Vy = Viy - gt
    4) X = Xi + Vixt = (ViCos(theta)) t



    3. The attempt at a solution

    B/c the ball is shot at 1.7m off the ground I reduced this to zero thereby reducing the height of the hoop by 1.7m.

    So I have the following givens:
    x= 3.05m
    y = 3.05m - 1.7m = 1.35m
    Theta = 60 degrees
    Xi = 0 m
    Yi = 0 m
    Time is unknown
    Solving for Initial Velocity (Vi)
    I am assuming max height is enough to clear the hoop at 1.35m. So at max height Velocity ='s 0 m/s.

    So now I used X = Xi + Vixt = ViCos(Theta)t

    3.05 = 0 + Vi (Cos 60) t
    t = 6.1 / Vi

    Now I take t and plug into y = Yi + Viyt - 1/2gt^2
    1.35m = 0 + Visin(60)t -1/2(9.8)(t^2)
    1.35 = (0.866025Vi)(6.1/Vi) - 4.9 (6.1/Vi)^2
    1.35 = 5.282754 - 182.329/Vi^2
    -3.932754 = -182.329 / Vi^2
    Vi^2 = 46.36165903
    Vi = 6.8089 m / s

    Does that seem right or am I not taking into account other factors. Thanks.
     
    Last edited: Sep 21, 2009
  2. jcsd
  3. Sep 21, 2009 #2

    berkeman

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    I'm confused by this. Don't you mean to say Viy is 0, not Vmax?
     
  4. Sep 21, 2009 #3
    I believe that intial velocity is unknown since the ball is already traveling at some speed in m/s. However, at the max height of the ball when it reaches the hoop would have a velocity of 0 with a negative accelleration of gravity or -9.8m/s^2.

    I have revised my attempt at the problem above. If anyone can confirm or discount my work would be much appreciated.

    Thanks!
     
  5. Sep 21, 2009 #4

    berkeman

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    Ah, I understand now. You mean Vy(Ymax) = 0. Got it.
     
  6. Sep 21, 2009 #5

    berkeman

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    This step looks wrong:

    Did you drop the Vi term? The first line is a quadratic equation with both Vi and Vi^2 terms. You would use the quadratic formula to find the solution...
     
  7. Sep 21, 2009 #6
    Sorry I added another Vi.

    So
    1.35 = (0.866025Vi)(6.1/Vi) - 4.9(6.1/Vi)^2
    which then leads to...
    1.35 = 5.282754 - 182.329 / Vi^2

    So the Vi's cancel eachother out so I don't have to use the quadratic forumula. I wasn't sure about this step either cause it seemed weird that they cancelled out without having to use the quadratic formula to solve for Vi.

    Thanks.
     
  8. Sep 21, 2009 #7

    berkeman

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    Ah, got it. If you have a graphing calculator, you could plot the trajectory that you've calculated, to see if it does what you expect it to.
     
  9. Sep 21, 2009 #8
    Thanks for the help Berkeman. When I calculate out time and re-plug that back in with the inital velocity I calculated, I would get to the 1.35m (a.k.a. the hoop.)

    However, I was wondering if I can just reduce the height where the ball was initially shot from and then just subtract that out of the initial height of the hoop. Is that fine to do w/o changing any other value as far as giving me the same answer? Thanks.
     
  10. Sep 21, 2009 #9

    berkeman

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    I believe that is fine. All you are doing is setting the zero point for y to be the start of the ball's flight. You are drawing the origin of your x-y axes at the player's hand instead of at his feet. Good simplification.
     
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