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Physics 20-1 Final Review: Projectiles

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A student is standing on the top of a building and he throws an object into the air with a speed of 16 m/s at an angle of 25 degrees above the horizontal. If the building is 75 m tall, how far from the base of the building will the object hit the ground?

    2. Relevant equations

    v = d/t
    d = vit + 1/2at2

    3. The attempt at a solution

    i tried to break it up into horizontal and vertical components and break it down into speed, then distance, then time. but nothing i do seems to work.
     
  2. jcsd
  3. Jan 18, 2009 #2

    LowlyPion

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    What do you get for the time to max height?
     
  4. Jan 18, 2009 #3
    a = v2 - v1 / t
    t = v2 - v1/ a
    t = 0 - 16 / -9.81
    t = 1.63 s.
     
  5. Jan 18, 2009 #4

    LowlyPion

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    There's your problem right there.

    The vertical component of velocity is what must be taken to determine time to max height.

    That is 16*sin25.

    The horizontal component is 16*cos25.
     
  6. Jan 19, 2009 #5
    i still get the wrong answer....
     
  7. Jan 19, 2009 #6
    where are u setting your zero? the top of the building or the bottom?
     
  8. Jan 19, 2009 #7
    huh? if you mean velocity wise, at the top for vertical motion.
     
  9. Jan 19, 2009 #8

    LowlyPion

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    So what is your time to max height?

    You know that must be 16*Sin25/g = t

    And how high is that from launch? y = 1/2 g*t2

    Now it is a simple free fall problem to determine the time to hit the ground isn't it?

    With the total time ... and the horizontal component of velocity ... viola your distance along the ground.
     
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