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Glad to have this tool (another basketball problem)

  1. Apr 12, 2007 #1
    May I first of all prelude this by saying that Physics has been a huge challenge for me. So, baby steps please.

    1. The problem statement, all variables and given/known data
    2.05-m-tall basketball player takes a shot when he is 6.02 m from the basket. If the launch angle is 25 degrees and the ball was launched at the level of the player's head, what must be the release speed of the ball for the player to make the shot? The basket is 3.05M above the floor.

    2. Relevant equations

    This is an area where I struggle. If I were given the formule, I, of course, could plug in the necessary information. I think...

    Theta=25 degrees
    Yo=2.05m above ground.
    Y1=3.05m above ground
    Ynet=1m (Y1-Yo)
    t=? (sqrt 2y/g) (positive gravity because the ball is going up...right?):uhh:

    3. The attempt at a solution

    X=VXoT=Vo COS 25 degrees T = Vo(.99)T
    Y=VYoT-1/2 g(t)sqrd=Vo SIN 25 degrees = Vo(-.13)T - 1/2g(t)sqrd

    Y=Vo-.13T-1/2(9.8)(t)sqrd = Vo-.13T-4.9(t)sqrd

    This is as far as I can take it...obviously it can be taken further, but I'm not sure where or how to take it from here.
  2. jcsd
  3. Apr 12, 2007 #2


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    Two points, first it looks like you are finding the values of your trig functions at 25 radians, not 25 degrees. Second, if you put T=0 in your X and Y equations you'll see that you get X=0, Y=0. So they are written with the origin of the coordinate system at the players head. This is fine, but bearing this in mind - what final values of X and Y do you want to solve for?
  4. Apr 12, 2007 #3

    I'm sorry, you lost me. I hate to be blunt but what would be the next step. I have no idea.
  5. Apr 12, 2007 #4


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    First off, correct the values of the trig functions. They are wrong for the reason I've already mentioned. Second correct your Y equation to include the initial position of the players head eg Y=Y0+VYoT-1/2 g(t)sqrd. Now put X=X1 and Y=Y1 into these equations and solve for V0 and T.
  6. Apr 13, 2007 #5

    X=VXoT=Vo COS 25 degrees T = Vo(.91)T
    Y=VYoT-1/2 g(t)sqrd=Vo SIN 25 degrees = Vo.42)T - 1/2g(t)sqrd
    so then you're saying
    Y=Y0+VYoT-1/2 g(t)sqrd

    Okay...am I heading in the right direction? There are two unknowns..How do I take care of 2 unknowns?
  7. Apr 13, 2007 #6
    yes you are headed in the right direction. By two unknowns which ones are you referring to? Never mind I see Dick's post now.

    Bear with me, I'm confused a bit by the notation and need a moment so sort this out. I'll be back in a couple.
    Last edited: Apr 13, 2007
  8. Apr 13, 2007 #7
    ok. lets look these over again just so we are on same page,


    and Yf=Yi+Vo(sin25)*T+1/2aT^2 where a=-9.8 Note that Yf-Yi=1m

    I suggest substituting the quantity 6.02/(Vo*cos25) for T in the bottom eqn.

    Just a tip for future work--this applies here and on tests as well, try to avoid the temptation to do the arithmetic on the way. If you leave it in terms of the original quantities the logic is much easier to follow, and a small mistake in arithmetic will not cost you so many points.
    Last edited: Apr 13, 2007
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