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Freefall basketball problem

  1. Aug 14, 2015 #1
    Hello first time posting in the Physics forum!!

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

    A varsity player is attempting to make a shot. The ball leaves the hands of the player at an angle of 50 degrees to the horizontal at an elevation of 2 meters above the floor. The skillful player makes the shot with the ball travelling precisely through the center of the ring (8 meters from the player and 3 meters above the floor). To loud cheers, calculate the speed at which the ball left the hands of the player. Sketch the problem!

    2. Relevant equations
    Y=Vot-1/2g(t)^2
    Voy = Vosintheta
    Vox = Vocostheta

    3. The attempt at a solution
    ON0PJMy.png
    Vox = Vocostheta -> 8=Vocos(50degrees) -> Vo=12.446m/s
    Then I plug it to X=Vot+1/2at^2 a is always 0 right? So it becomes X=Vot -> X=12.446(t) -> t=(0.643s)
    Then I plug the time to the y equation of free fall.. Delta Y is 1 right? Because 3-2=1
    Y=Vot-1/2gt -> 1=Vo(0.643)-1/2(9.8)(0.643)^2 Vo=4.7059m/s
    y
    I'm not sure if my answer is right.. if not can someone guide me....
    If it is wrong sorry If I got a wrong answer.. This is why I post in this forums.
     
  2. jcsd
  3. Aug 14, 2015 #2

    BvU

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    Hello VZ, welcome to PF :smile: !

    Always a good idea to include dimensions in your calculations -- and to check them. In your case, you go off the rails in the very first line (sorry to say):

    ## 8 \;{\rm m} = V_{0,x} = V_0 \cos\theta \;{\rm m/s}## can never be right !

    Thing to do is write expressions for x(t) = 8 m and y(t) = 3 m, and eliminate t. Best to work with symbols until you have an expression for ##\theta##. Then check the dimensions and do the calculator work.

    Oh, and: strange you should ask if your answer is right: I see two answers for v0
     
  4. Aug 14, 2015 #3
    I don't know what you mean by dimension buuuuuuut..
    ΔX = Vo(t)
    ΔY = Vo(t)-1/2(g)(t)^2

    Do you mean that I should experiment with this two formulas? To get Vo? or to get t?
     
  5. Aug 14, 2015 #4
    First - you have equated horizontal velocity to horizontal displacement . This is wrong .
    So your second and third equations automatically become wrong .
     
  6. Aug 14, 2015 #5
    Yeap I realized that after sir BvU post. Sooooo if it is okay can you guide me in this question?
     
  7. Aug 14, 2015 #6
    Consider two unknowns v and t . Obviously v represents velocity of object , and t the time taken to reach the hoop .

    What you'll need to do is form two equations . Use the fact that horizontal velocity of object is v*cos(θ) and vertical v*sin(θ) .

    Can you manage the two equations ?
     
  8. Aug 14, 2015 #7
    Okay will do! I'll update you if I'm done. Thank you
     
  9. Aug 14, 2015 #8
    SIR I THINK I GOT IT. Okay here goes

    ΔX = Vox(t) right? bc a is always 0
    ΔX/Vox = t but we know that Vox=VoCos50 and ΔX = 8
    so 8/Vocos50 = t
    Voy=Vosin50 right?
    So
    Δy=Voy(t)-1/2g(t)^2
    3-2=(Vosin50)(Vocos50)-1/2(9.8)(Vocos50)^2
     
  10. Aug 14, 2015 #9
    You need to substitute correctly for t .
     
  11. Aug 14, 2015 #10
    typo 8/Vocos50 but it is right the context is still there..
     
  12. Aug 14, 2015 #11
    Yes it is .
     
  13. Aug 14, 2015 #12

    BvU

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    No experimenting: you want to manipulate these two equations with two unknowns in such a way that you get one equation with one unknown, namely ##\theta##

    The way you write them is confusing -- one Vo with two different meanings, and Vo(t) makes it look as if Vo is a function of t --$$
    \Delta X = v_0 \cos\theta\; t \\ \Delta Y = v_0 \sin\theta\; t - {1\over 2} gt^2 $$ is a lot clearer. See also here

    [edit] oops, I'm lagging. This is in response to post #3 -- a bit late.
     
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