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VERY DIFFICULT 2-D Motion Problems!

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    1. A baseball player hits a homerun and the ball is caught by a person in the stands. It is caught 7.50m above the point it was hit. At the moment it was caught, it has a velocity of 36.0m/s at an angle of 28 degrees below the horizontal. What was the initial velocity of the ball when it was hit.

    2. Basketabll hoop is 3.05 m above the playing surface. A basket is made. The ball reached a max. height that was 2.00m above the height of the basket hoop. The basketball was launched from a height of 1.95m. If the ball traveled a horizontal distance of 5.20 m in 2.00 seconds, what was the initial velocity of the basketball?



    2. Relevant equations
    v f = vi + at
    delta x= vi + 1/2 a T^2
    vf ^2= Vi^2+ 2 a delta x


    3. The attempt at a solution
    I have no idea where to start. Can someone lead me down the right path to answering these questions?
     
  2. jcsd
  3. Oct 21, 2009 #2

    cepheid

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    For number 1, you can calculate the vertical component of the ball's velocity at the moment it was caught, and from that can easily deduce from what maximum height it began to descend. The maximum height reached tells you the initial vertical component of the velocity (upon being hit).
     
  4. Oct 21, 2009 #3

    cepheid

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    For number 2, the time taken to travel a horizontal range of 5.20 m tells you the initial horizontal component of the velocity. The max height reached tells you the vertical component of the initial velocity.
     
  5. Oct 21, 2009 #4
    I don't understand what you mean by that. Can you please explain it a bit further?
     
  6. Oct 21, 2009 #5

    cepheid

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    In problem 1, you know the magnitude and direction of the velocity vector of the ball at the moment it is caught. Therefore, you can resolve the velocity into horizontal and vertical components. The vertical component (how fast it was travelling "downwards") tells you what maximum height it fell from (measured above the position at which it was caught), because you know that it was accelerating under gravity.
     
  7. Oct 21, 2009 #6
    Vyo=Vosin(theta)
    Vxo=Vocos(theta)

    Right down all your knows in a t chart. one side you x's and the other your y's

    ay=-9.8
    delta y=7.50m
    Vo=36m/s
    There are some more
    x side equation is(Delta)x=VoT
    Y equation (Delta)y=Vyot+1/2Ayt^2

    Solve for time then you can plug that in and get Vyo
     
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