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Velocity equation help

  1. Oct 12, 2007 #1


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    You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. V Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.


    What minimum speed will allow the ball to clear the roof?

    okay i figured out this much so far:
    delta x= 9m
    9= (Vo cos theta)T
    delta y=5
    5= (Vo sin theta)T- .5gt^2

    Vy=Vosintheta - gt=0

    t=Vosintheta / g

    and now i know i am suppose to plug in T into the x equation and y equation to solve for Vo, but i cant figure out to do that?
  2. jcsd
  3. Oct 12, 2007 #2


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    like i know if you substitute t in the first equation you can get,

    9= (Vo cos theta) (Vosintheta / g) so does that equal
    9= (2Vo costheta sin theta) / g

    and then

    5= (Vo sin theta) (Vosintheta / g) - .5g (Vosintheta / g)^2
    5= Vosin2theta/2g

    but then what?
  4. Oct 12, 2007 #3


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    please anyone?
  5. Oct 13, 2007 #4


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    if anyone could please look at my work that would be great. i am online to discuss. please.
  6. Oct 13, 2007 #5


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    thank you first of all for responding! but i am very confused as what i am suppose to do? i know this much, that t=Vosintheta / g, but where am i suppose to substitute that into?
  7. Oct 13, 2007 #6
    See here you know the maximum height and that is 3 + 3tan(45).

    Now the range is 18 m.

    Hence we have two variables and two equations.

    So simultaneously solve the following two.

    [tex]H_{max}= \frac{v^2sin^2\theta}{g}[/tex]

    [tex]R = \frac{v^2\sin2\theta}{g}[/tex]
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