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A boy throws a ball

  1. Sep 22, 2009 #1
    1. A boy 11.0m above the ground in a tree throws a ball for his dog, who is standing
    right below the tree and starts running the instant the ball is thrown.
    If the boy throws the ball upward at 40.0° above the horizontal, at 7.50m/s .
    What is velocity (m/s)? and distance (m)?


    2. y=yo+voy*t-1/2gt^2

    y= yo+sin[tex]\theta[/tex]*t-1/2gt^2

    1/2gt/yo= voy

    R= Voy*t=Vosin[tex]\theta[/tex]*t

    t= [tex]\sqrt{}[/tex](2y/g)^2

    3. I plugged the known variables into the equations i listed above. First for time, I got t=1.5s. Then I used the angle and time found to find distance, R=(7.5)(sin40)=4.82 m. Which doesn't make sense to me. Any help is appreciated. Thanks!

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 22, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Kyle! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    How did you get t= √(2y/g)^2 from that? :confused:
     
  4. Sep 22, 2009 #3
    t= √(2y/g)^2 is just a derivation of some sort that my professor had given us to figure time in 2d motion.
     
  5. Sep 22, 2009 #4
    Would t=Vo/-a work from the base equation Vy=Vo+at?
     
  6. Sep 23, 2009 #5

    tiny-tim

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    That only works if v0 = 0 … just look at the original equation.

    Stop looking for shortcuts!
     
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