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Need Help! Launches at angles

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data
    a physics student pitches a sack of doorknobs into the air with an initial velocity of 24.5 m/s at 36.9degrees from the horizontal.
    i know the:
    time=3.0s
    horizontal distance traveled=58.8m
    time when sack is at max height=1.5s
    max height sack reaches=11.03m

    but i need to know how to figure:
    the velocity of the sack as it hits the ground=?

    ok so i have the correct answer its:
    24.5 m/s, trajectory is 322.9degrees or 36.9degrees with horizontal


    2. Relevant equations

    i need to know how you got this answer
    like what equations did you use?

    3. The attempt at a solution

    i got that far as to figuring out all the questions but how do you get that one???


    (my teacher gives us the question and the answer and wants to see our work...thats how i know that last answer)
     
  2. jcsd
  3. Sep 23, 2008 #2

    LowlyPion

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    What are the x and y components of velocity when the sack lands?

    [tex]\vec{V} = \vec{V_x} +\vec{V_y}[/tex]
     
  4. Sep 23, 2008 #3
    x=58.8 and y=0
    is that what your asking for??
     
  5. Sep 23, 2008 #4
    noo
    ok but how do you get the Vx and Vy
    the Vo=24.5
     
  6. Sep 23, 2008 #5

    LowlyPion

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    What was it horizontal velocity at that time + what was the vertical velocity.

    Add them together as vectors. You will get a magnitude and an angle.
     
  7. Sep 23, 2008 #6

    LowlyPion

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    You had to have had the horizontal velocity to know how far it traveled.
     
  8. Sep 23, 2008 #7

    LowlyPion

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    Vy is given by how fast it was going falling from 11 meters.

    Or Vy = g*t to fall from that height where t is the same t you found to get to that height.
     
  9. Sep 23, 2008 #8
    oh ok i have the initial horizontal velocity its 19.59 and the initial vertical velocity is 14.7

    do need any final velocitys to solve this?
    or am i going off the initial the whole time?
     
  10. Sep 23, 2008 #9

    LowlyPion

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    The initial Vx will remain the same.

    The final Vy will be downward. Since it was at the same height at the end that it started, downward velocity will be the same magnitude as the start, but opposite sign. But note it's only because it's at the same height.
     
  11. Sep 23, 2008 #10
    oh ok.

    ok so what like equatitions with the numbers in them would you use to got to the final answer of 24.5m/s?
     
  12. Sep 23, 2008 #11

    LowlyPion

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    The components form two sides of a right triangle. Any way you can think of to calculate the Hippopotamus?
     
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