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Basketball player shooting

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A basketball player is 15ft horizontally from the centre of the basket which is 10ft off the ground At what angle should the player aim the ball from a height of 8.2ft with a speed of 26fts^-1?


    3. The attempt at a solution

    [itex] x = vi cos . t [/itex]
    [itex]t = \frac{x}{vi cos Θ}[/itex]

    sub t into y(t):

    1.8 = [vi^2 sin2Θ - gx^2]/2vi^2 cos^2 Θ

    Θ= 22.6

    The answer stated solution to be 66°
     
  2. jcsd
  3. Jan 29, 2014 #2

    BvU

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    Hello again. Don't recognize your y(t) expression. Where did you start from and how did you derive it ?
     
  4. Jan 29, 2014 #3

    Y(t) is the y displacement as a function of time.
    It has the form yf = yi + vyi. t - 0.5gt^2
    t= x/ vi cos (theta), vi=26ft/s, yf=10, yi=8.2 and g = 32ft/s^2
    I sub t and the known variables into the above equation. What I got was 22 degrees. But this contrasts with the book's
     
  5. Jan 29, 2014 #4

    BvU

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    Funny, I substitute t = x/(vi cos(theta)) in vyi * t with vyi = vi sin(theta) and get x * sin(theta)/cos(theta) !? You can change to 2theta which should introduce a 2 (which you do take into account) but it shouldn't let the x disappear !
     
  6. Jan 29, 2014 #5
    Where is your vi? Shouldn't it be x.vi sinΘ/vi cos Θ?
     
  7. Jan 29, 2014 #6

    BvU

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    vi/vi = 1
     
  8. Jan 29, 2014 #7
    Capture.JPG

    The answer doesn't tally with the book's
     
  9. Jan 29, 2014 #8

    BvU

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    I see you multiply by vi^2cos^2(theta) on the righthand side. To keep the equality valid, you have to do so also on the lefthand side. I.e. the 1.8 changes...
     
  10. Jan 29, 2014 #9

    haruspex

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    Quite so, but I don't think this approach is leading anywhere.
    negation, go back to the third line of your attachment: "1.8 = " etc.
    sin/cos = tan; 1/cos2 = sec2.
    Using sec2 = 1 + tan2 you get a quadratic in tan.
     
  11. Jan 29, 2014 #10

    I overlooked that but still it's not working out
     
  12. Jan 29, 2014 #11

    Let me try
     
  13. Jan 29, 2014 #12
    I'm getting 22.4 degrees. Still, it doesn't tally with the book's
     
  14. Jan 29, 2014 #13

    haruspex

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    I get a much larger angle. Please post your working.
     
  15. Jan 30, 2014 #14
    Capture.JPG
     
  16. Jan 30, 2014 #15

    haruspex

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  17. Jan 30, 2014 #16

    Ought to have been a negative
     
  18. Jan 30, 2014 #17
    Did you got 72.8?
     
  19. Jan 30, 2014 #18

    BvU

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    No, should be 65.67 degrees if g = 32
    Filling in 72.8 doesn't satisfy your final equation with the corrected sign: some 9630 off!
    A check you should always do (if you have time enough)
     
  20. Jan 30, 2014 #19
    Strange. I'm not getting the answer even after correcting that positive sign and changing it to a negative.
    But I think it's good enough to leave it here since the problem solving part is much more important than the minute calculation details.
     
  21. Jan 30, 2014 #20
    solved.
     
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