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Projectile motion

  1. Jun 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A bastketball player is 2m tall and stands 10m from the basket.The height of the basket is 3m and the player throws the ball with speed 11m/s. At what angle with respect to the horizontal should he throw the ball so that is goes through to basket?

    2. Relevant equations
    g=gravity, h= maximum height vi= initial velocity x=angle R= maximum range
    h=vi^2 * (sin(x))^2 / 2g
    R= vi^2 * sin 2x /g

    3. The attempt at a solution
    Here is what've done:

    h=vi^2 * (sin(x))^2 / 2g
    x= Inverse SIN (SQRT ( h*2g / vi^2 ))
    height = 1m
    x= Inverse Sin(SQRT (1*2(9.8)/ 11^2)) = 23.73 degree

    R= vi^2 * sin 2x /g
    x= (Inverse SIN(R*g / vi^2)) / 2
    R= 10m
    x= (Inverse SIN(10*9.8 / 11^2)) / 2 = 27.04degree

    both angles aint the same,
    is it because the height of the basket is different to the player's height?
    do i need to put "h=vi^2 * (sin(x))^2 / 2g " and
    "R= vi^2 * sin 2x /g " into one eqn in order to get the angle?
     
    Last edited: Jun 4, 2008
  2. jcsd
  3. Jun 4, 2008 #2

    Kurdt

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    You need to resolve the problem in to horizontal and vertical components. What is the equation describing the horizontal motion of the ball? what is the equation tha describes the vertical motion of the ball?
     
  4. Jun 4, 2008 #3

    tiny-tim

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    Hi hayowazzup! :smile:

    Doom … despair … calamity … :cry:

    Those equations won't help.

    You don't need the maximum height, and you don't need the range (which, as you say, is only for heights which are the same).

    You'll have to start from scratch, and work out your own equation. :smile:

    Hint: deal with the x and y components separately … oh, and don't call the angle x as well, that'll only confuse you … have a theta …
     
  5. Jun 4, 2008 #4
    thnx
    so,
    cos θ=Vxi / Vi => Vxi=cos θ * Vi
    sin θ=Vyi/ Vi => Vyi=sin θ * Vi
     
  6. Jun 4, 2008 #5

    tiny-tim

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    That's right! :smile:

    So find out from Vxi how long it takes to go 10m horizontally.

    Then concentrate on Vyi, and work out a formula to tell you what its height is at that time.

    (And remember that the height has to be +1m for the second time … the ball has to go through from above!!). :smile:
     
  7. Jun 5, 2008 #6
    ∆x = Vx0 * t
    t= ∆x / Vx0
    t= 10m / Vx0
    but how do i find out the initial horizontal velocity??
     
  8. Jun 5, 2008 #7

    tiny-tim

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    That's right! :smile:

    So t = 10/11cosθ.

    Vy0 = 11sinθ, so what is the height at t = 11cosθ? :smile:
     
  9. Jun 5, 2008 #8
    do you mean the Vyf?
    0m?
     
    Last edited: Jun 5, 2008
  10. Jun 5, 2008 #9

    tiny-tim

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    I'm confused. :confused:

    Vyf is a velocity.

    Hint: ay = -9.8. Vy0 = 11sinθ. So what is the formula for y (as a function of t)? :smile:
     
  11. Jun 5, 2008 #10
    o i see,
    ∆y =Vy0 * t - (1/2)g t^2
    ∆y = 11sinθ * t - (1/2) *-9.8 * t^2
     
  12. Jun 5, 2008 #11

    tiny-tim

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    That's right! :smile:

    So the height at t = 11cosθ is … ? :smile:
     
  13. Jun 5, 2008 #12
    ∆y = 11sinθ * (11cosθ) - (1/2) *-9.8 * (11cosθ) ^2
    ∆y = 121 sinθ *cosθ + 592.9 (cosθ)^2
     
  14. Jun 5, 2008 #13

    tiny-tim

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    hmm … I've just noticed there's too many minuses in there. :frown:
    ok … now put ∆y = 1, and solve using usual trig methods. :smile:
     
  15. Jun 5, 2008 #14
    unfortunately, i don't have a clue about how to solve the equation..
    can you teach me how to do it , or do you have any references that I can look at?thnx
    I guess the first step is that I need to factorise it
    1 = cos θ(121 sinθ + 592.9 (cos θ))
    1/cos θ = 121 sinθ + 592.9 (cos θ)
     
  16. Jun 6, 2008 #15

    Kurdt

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    Something has gone slightly wrong somewhere. There should be 1/cos θ's hanging around.
     
  17. Jun 6, 2008 #16
    do you mean shoudnt?
    should i use the rule sin² + cos²=1
    but then how do i get rid of sinθ *cosθ ?
     
  18. Jun 6, 2008 #17

    Kurdt

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    Nope I mean you worked out the time with the x-component of the motion and that came out as t = 10/11 cos θ.
     
  19. Jun 6, 2008 #18
    1 = 11sinθ (10/11cos²θ) + 4.9 (10/11cos²θ)
     
  20. Jun 6, 2008 #19

    tiny-tim

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    Hi hayowazzup! :smile:

    (btw, it's 1 = 121 sinθ *cosθ - 592.9 (cos²θ), with a minus in the middle)

    The way I'd solve this … and I suspect it's not the quickest :rolleyes: … is to rewrite it in the form A.cos2θ + B.sin2θ = C , then define tan φ = A/B, and solve for (θ - φ). :smile:
     
  21. Jun 6, 2008 #20
    hi, right so,
    sin 2θ= 2sinθcosθ
    cos 2θ= 2cos²θ -1

    1 = 121 sinθ *cosθ - 592.9 (cos²θ)
    0 = 60.5 (2sinθcosθ) - 296.45 * 2(cos²θ) - 1
    0 = 60.5 sin 2θ - 296.45 cos 2θ
    tan φ= -296.45 / 60.5 = -4.9
    φ = tan-1 (-4.9) = -78.46°
     
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