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Vertical Displacement

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi guys, I am working through a very long problem and I am stuck at this part.

    So I have a 70kg human being launched at a 90 degree angle at 15mph. I need to figure out how high in the air he will get before velocity reaches zero again and he begins to fall.



    2. Relevant equations
    Vf = Vi + at

    Rearranged to: a = (Vf-Vi)/ t

    A- time
    Vi- initial velocity
    Vf- final velocity
    t- -9.8m/s^2


    3. The attempt at a solution

    After I rearranged the problem I got a = (0-2.5)/-9.8

    a = .255

    Now I used D= Vi*t (1/2) a t ^2

    6.25 (.255) = 1.59

    1.59 + 1/2 (-9.81)(.255)= 1.27 m


    Is this correct??
     
  2. jcsd
  3. Mar 2, 2015 #2

    SteamKing

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    Not even close.

    Since the human is being launched into the air (presumably without a rocket attached to his back side), he is not accelerating off the ground, but has a constant velocity of 15 mph as he leaves the ground. The only thing retarding his upward movement is the gravitational pull of earth.

    What is g for earth in foot-second units?

    BTW: I've moved this thread to the Intro Physics HW forum, which is more appropriate given the nature of the problem.
     
  4. Mar 2, 2015 #3
    It would be 9.8m/s^2

    So, -32.15f/s^2 right?

    Now I convert 15mph to 22 f/s.

    Now I multiply 22f/s x 32.1 f/s^2 = 706 f/s^2

    Is this right?
     
  5. Mar 2, 2015 #4

    SteamKing

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    You're going from bad to worse.

    Why are you multiplying the acceleration due to gravity by the initial velocity of the human?

    For the record, the units of ft / s multiplied by ft / s2 are not equal to ft / s2.
     
  6. Mar 2, 2015 #5
    I am really trying hard to work through this one. Could you suggest a direction I should take?

    Perhaps I should not be multiplying at all. Since the acceleration due to gravity is 9.8ms^2 I should subtract that from the 22f/s since the gravity is "retarding" the human. Am I getting anywhere here?
     
  7. Mar 2, 2015 #6

    SteamKing

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    You're not getting anywhere because you aren't using the correct equation to solve this problem. You want to select the kinematic equation or equations which will allow you to calculate the distance traveled by the human after he is launched.

    http://www.ronknott.com/MEI/MechSuvatEquns.html
     
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