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Physics of basketball

  1. Jan 24, 2012 #1
    Hello,

    I am designing a robot that shoots basketballs of diameter 7 in and am trying to figure out a couple things.

    First:

    I was talking to a friend and they were saying that when you shoot a ball energy gets lopped off as it is shot.

    I'm wondering if this is energy lost due to having to spin the ball or energy lost due to air resistance.

    also I am trying to figure out the angle we have to shoot our ball given the velocity, x position and y position from the hoop.

    what I have done so far is use x final (xf) = x initial (xi) + V initial in x direction (Vix) *t

    and yf = yo + vyo*t - 1/2 gt^2

    so say the hoop is 3 meters away in x direction,
    my shooter is on the ground at x = 0
    the hoop is at y= 1.5 meters
    the velocity of the ball coming out of the shooter is v= 9.5 m/s

    so basically i would use x equation first

    xf = x0 + vox *t

    3 = 0 + 9.5 cos theta *t

    solve for t

    t = 3 / 9.5 cos theta


    then use y equation

    yf = yo + voy *t - 1/2 g t^2

    1.5 = 0 + vo sin theta * (3/9.5 cos theta) - 4.9 (3/9.5 cos theta)^2

    I don't know how to solve this though....

    Or is there a better way to solve this?

    Thanks
     
  2. jcsd
  3. Jan 25, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    I haven't looked at your mathematics; just wanted to address this.

    Sometimes, especially when you don't need a highly precise answer, it is feasible to step through all possibilities, and use the one that comes closest to the solution. So here you could step through all thetas from say, 5 deg to 85 deg, in steps of 1 deg, or even 0.5 deg. It sounds like the application is one where time is not at a premium.
     
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