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Homework Help: Another Projectile Motion

  1. Oct 15, 2005 #1
    Basketball player is 2m tall. He shoots at 40 degrees over a distance of 10m into a basketball net that is 3.05m. Whats the initial velocity.

    Theta=40 degrees

    Vxi = Vi cos(40) Ax=0 xi=0 xf=10
    Vyi= Vi sin(40) Ay=-9.80 yf=1.05 (difference between player and net)

    1. First I wanted to solve for time

    Xf = Xi + Vxi*t + (1/2)Ax*(t^2)
    10 = 0 + (Vi*cos(40))t + (1/2)(0) (t^2)
    10 = (Vi*cos(40))t

    t = (Vi*cos(40))/ 10

    2. Then, I'd use the Yf formula to plug my new time into to get Vi since Vyi= Vi sin(40)

    Yf = Yi + Vyi*t + (1/2)Ay*(t^2)
    1.05=0 + (Vi*sin(40))t - (1/2)(-9.80)(t^2)

    ---then substituting t into the 1.05=0 + (Vi*sin(40))t + (1/2)(-9.80)(t^2)
    and after all the algebraic dust settled I got ---

    Vi = 7.159.... the problem is the book says 10.7 m/s. Where did I go wrong??
    Last edited: Oct 15, 2005
  2. jcsd
  3. Oct 15, 2005 #2


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    Homework Helper

    You have used too many "-"-s.
  4. Oct 15, 2005 #3


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    For the time, it should be,

    t = 10/(Vi*cos(40))
  5. Oct 15, 2005 #4
    that was a typo
  6. Oct 15, 2005 #5
    Holy crap... i'm staring at my page of work and you are sooooo right. Let me try that and see if it works now.:smile:
  7. Oct 15, 2005 #6
    It worked... It worked!!!

    That was the fix. I had the equation for t wrong. I got the 10.7 book answer now. Thank you sooooooooooo much!!!
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