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Basketball problem

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A basketball player who is 2.00 m tall is standing on the floor L = 8.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 30.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.


    2. Relevant equations

    y=Yo+Voyt-(g/2)t squared
    X=Xo+Voxt-(g/2)t squared


    3. The attempt at a solution

    Vxo=cos(30)
    Vyo=sin(30)
    Vo (cos 30)=8

    3.05=2.00+Vo-4.9t squared
    3.05=10-4.9t squared
    t=square root (6.95/4.9)
    =1.191

    Vo=8/(.7071*1.191)
    =9.499


    However, this is wrong...any suggesions??
     
  2. jcsd
  3. Feb 1, 2007 #2
    first check your cos 30 value.
     
  4. Feb 1, 2007 #3
    oops...the the cosign value of 30 is .8660254038. So I take that value and multiply it by the square root of 6.95/4.9. Which is 1.191. I divide 8 by that product. Correct? My answer comes out to be 7.7561749; however, that is appearantly wrong. What am I doing wrong?
     
  5. Feb 1, 2007 #4
    your sin value gets bollixed as well and the Vot (first order term) disappears--2Vo becomes 10. There may be more issues but start with those. Careful algebra for starts.
    John
     
  6. Feb 1, 2007 #5
    I have 2 Vo becoming 10 in my equation. I have:
    3.05=10-4.9t squared
     
  7. Feb 1, 2007 #6
    think we'are cross threaded, maybe I'm missing something
    but,

    y=Yo+sin(30)Vo*t+1/2at^2 going to 3.05=10-4.9t squared has me confused
     
  8. Feb 1, 2007 #7
    bc 3.05 was the height of the basket. I set the equation :Yo+sin(30)Vo*t+1/2at^2 equal to the height of the basket. I am pretty confused. I'm not sure what I'm doing at this point:uhh: I have my final equation looking like Vo=8/(cosign30 * 1.191). I found the 1.191 by taking the square root of 6.95/4.9.......AHHH!!!
     
  9. Feb 1, 2007 #8
    Lets try a fresh start.

    first assuming a conventional x/y horizontal and vertical coordinate system:

    your eqn for Y is correct, x is not. Since there is no acceleration in in this direction, its simply
    x=cos(30)Vo*t. But we do know how far the ball has to travel 8m.
    so 8/cos(30)=Vo*t Now can we use that info in the Y eqn to make life simpler? For sure and I think you were close, so patience, my friend. Beware tho i think there may still be one wrinkle ahead.
     
    Last edited: Feb 1, 2007
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