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Inclined throw

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data
    First i apologize for my english, i don't know the customary names.
    A baseball player hits the ball at an angle of 300 above the horizon. it lands 100 meters away at the same height (another player catches it).
    What is the velocity at the beginning?

    2. Relevant equations
    Constant gravitational acceleration: V=V0-gt

    3. The attempt at a solution
    Total time: the time needed for going up and down (i separate the movements to horizontal and vertical):
    [tex]0=V_0\sin30^0-gt \rightarrow t_{tot}=gV_0[/tex]
    Horizontal movement:
    [tex]100=V_0\cos30^0\cdot t=0.87V_0\cdot 10V_0[/tex]
    [tex]\rightarrow 100=8.7V_0 \rightarrow V_0=3.4[/tex]
    The result should be 10 times faster: 34 m/sec
     
  2. jcsd
  3. Apr 30, 2014 #2
    How did you go from:
    [tex]0=V_0\sin30^0-gt[/tex]

    to the following:
    [tex]\rightarrow t_{tot}=gV_0[/tex]

    for the vertical component?
     
  4. Apr 30, 2014 #3

    BvU

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    is strange. How do you do that and what do you calculate? Your relevant formula would say: that is where V = 0 in the vertical direction.
     
  5. Apr 30, 2014 #4
    Yes, the time till V=0, double, is the time to go up and down again
     
  6. Apr 30, 2014 #5

    BvU

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    Good. It is less confusing if you actually mention that extra step.
    Now if V0 - gt = 0 then V0 = gt so what is T total ?
     
  7. Apr 30, 2014 #6
    Thanks very much, i solved it thanks to Goraemon!!
     
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