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Projectile Motion of launched ball

  1. Sep 17, 2008 #1
    Projectile Motion - Finding The Angle

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

    You are standing 10 m away from a basketball net that is 14 m above the ground. Assume delta d[y] is 14 m, do not account for your height. You launch the ball with a velocity of 25 m/s and it goes in the hoop.

    At what angle(s) could you have thrown the ball so that it goes through?


    2. Relevant equations
    The kinematics equations and others that I may not be aware of.
    [tex]\vec{v}=25 m/s[/tex]
    [tex]\Delta\vec{d}[x]=10 m[/tex]
    [tex]\Delta\vec{d}[y]=14 m[/tex]
    [tex]\vec{a}[y]=-9.8 m/s^2[/tex]
    [tex]\theta = ?[/tex]
     
    Last edited: Sep 17, 2008
  2. jcsd
  3. Sep 17, 2008 #2
    Try using the equations of projectile motion:

    vx=vCos(theta)
    vy=vSin(theta)+0.5at

    x=x0+vx0t
    y=y0+vy0t+0.5at2
     
  4. Sep 17, 2008 #3


    That's exactly what I did but I don't know where to go from there.
     
  5. Sep 17, 2008 #4
    eliminate t with algebra and then solve for theta.
     
  6. Sep 17, 2008 #5
    Ok now I have all the equations I need:

    [tex]V[x]^2 + V[y]^2 = 25^2 = 625[/tex]
    [tex]14 = V[y] \Delta t - 4.9 t^2[/tex]
    [tex]10 = V[x] \Delta t[/tex]

    I know how to do simultaneous equations. I have to solve for V[x] and V[y] from the second and third equations and plug that into the first one to get time. Then I can plug time into any equation and solve for V[x] and V[y]. Once I have those, the angle is simply Tan-1(V[y]/V[x]).

    I did it all in mathematica (math software), but when I tried to actually solve it on paper, I scribbled 6 pages and couldn't solve it. The answer was right when done on the computer but now I have to do it algebraically. Can you help me solve it algebraically?

    So,

    [tex]V[x] = 10/t[/tex]
    [tex]V[y] = \frac{4.9t^2 + 14}{t}[/tex]

    Then,

    [tex]25^2 = V[x]^2 + V[y]^2[/tex]
    [tex]25^2 = (10/t)^2 + (\frac{4.9t^2 + 14}{t})^2[/tex]

    This is where it gets messy.
     
  7. Sep 17, 2008 #6
    What are the answers that mathematica gave you? I would like to see if mine are close after doing it on paper.
     
  8. Sep 17, 2008 #7
    I put in: Solve[25^2 == (10/t)^2 + ((14 + 4.9 t^2)/(t))^2, t]

    It gave me: {{t -> -4.43739}, {t -> -0.791265}, {t -> 0.791265}, {t -> 4.43739}}

    These two are the right answers for t: {t -> 0.791265}, {t -> 4.43739}

    All I want to know is how to get to this point on paper.
     
    Last edited: Sep 17, 2008
  9. Sep 17, 2008 #8
    your method of finding theta by first finding the value of t is horribly inefficient.

    Start with the equations:

    x=x0+vx0t
    y=y0+vy0t+0.5at2

    We know that at some time t the ball will have the coordinates (10, 14) which translates to:

    10=x0+vx0t
    14=y0+vy0t+0.5at2

    We assume the ball is launched from the origin so the equations simplify to:

    10=vx0t
    14=vy0t+0.5at2

    We are given that the magnitude of the balls velocity when it is launched is 25. That means:

    vx0=25Cos(theta)
    vy0=25Sin(theta)

    The equations now simplify to:

    10=25Cos(theta)t
    14=25Sin(theta)t+0.5at2

    The first equation can be solved for t and then subsituted into the second equation. After some shuffling and applying trig identities you arrive at the quadratic formula:

    0=-(14+0.08a)+10Tan(theta)-(0.08a)Tan2(theta)

    Substitute u = Tan(theta) into the expression and solve for u. You should get two answers. Use these answers in the equation Tan-1(u)=theta and you should get what you were looking for.
     
  10. Sep 17, 2008 #9
    Well sorrrrryyyyyy!

    Can you explain how you got from the top two equations to the quadratic equations? I tried it but I'm getting imaginary numbers.
     
  11. Sep 17, 2008 #10
    According to your equation, I did this:
    Solve[0 == -(14 + 0.08 (-9.81)) + 10 u - 0.08 (-9.81) u^2, u]

    It spits out: {{u -> -13.9493}, {u -> 1.20716}}

    So far, I have 3 different sets of answers (6 answers in total) which all seem to work out. There should be only 2 solutions. Which one is right?
     
  12. Sep 17, 2008 #11
    I didn't mean to be offensive. I just wanted to point out that there was a much easier method.

    Solving for t in the first equation gives:
    t=0.08Sec(theta)

    Inputting this into the second equations gives:

    14=25Sin(theta)[0.4Sec(theta)]-0.5a[0.4Sec(theta)]2
    14=10Tan(theta)-0.08aSec2(theta)

    There is a trig identity that says Sec2(theta) = Tan2(theta)+1 so

    14=10Tan(theta)-0.08a[Tan2(theta)+1]
    0=-(14+0.08a)+10Tan(theta)-0.08aTan2(theta)

    a=9.8
     
  13. Sep 17, 2008 #12
    a=9.80 not -9.80
     
  14. Sep 17, 2008 #13
    Thank you! It works now. Why is it not negative?

    And thanks again for all your help! It's clear now!
     
  15. Sep 17, 2008 #14
    My equation already takes into account that the acceleration due to gravity is pointing down, so all that is necessary is putting in the magnitude of the acceleration which is 9.80

    Do you have an answer key or something for this problem?
     
    Last edited: Sep 17, 2008
  16. Sep 17, 2008 #15
    No, but I know your equation is right, I simulated it.
     
  17. Sep 17, 2008 #16
    It is good to know it works and that I have not been wasting your time with wrong answers!

    :approve:
     
  18. Sep 17, 2008 #17
    And where exactly did you modify it so it takes into account that gravity is pointing down?

    Edit, nevermind I see it. Thanks so much!
     
  19. Sep 17, 2008 #18
    No problem at all.

    night :zzz:
     
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