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Dynamics Question

  1. Dec 4, 2007 #1
    1. [​IMG]
    [​IMG]




    2. d = 1/2at^2 + v1t

    v =d/t




    3.
    horizontal time it takes equals vertical time

    Horizontal
    t = d/v = 50 /(v1cos10)


    Vertical
    d = 1/2at^2 + v1t
    -0.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)

    i get v1 = 36.6 m/s

    -1.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)
    i get v1 = 34.8 m/s


    So my range is 34.8 m/s to 36.6 m/s

    for b part

    dv = v1(t) - 1/2(9.81)(t^2)

    maximum height occurs at half the time so i used 1/2(50/v1cos10)

    dv = 34.8(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/34.8cos10))^2
    i got 22.8 m

    for 2nd velocity

    dv = 36.64(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/36.64cos10))^2
    i got 23.0 m


    So my maximum height range is 22.8 m to 23.0 m

    According to my online assignment this is wrong

    Edit:
    i just noticed for maximum height i guess I also add 2.1 m if they are measuring from the Ground so 22.8 + 2.1 = 24.9 m and 23.0 + 2.1 = 25.1 m but these values still are somehow wrong........


    plz helppppppppppp

     
  2. jcsd
  3. Dec 13, 2007 #2
    [itex] y = v_{oy} t - \frac{1}{2} g t^2 + y_o [/itex] (1)

    [itex] y_{max} [/itex] occurs when [itex] y' = 0 [/itex]

    [tex] y' = v_{oy} - g t \longrightarrow t = v_{oy} / g[/tex]

    Substitute t in eq.(1)

    [tex]y = \frac{v^2_{oy}}{g} - \frac{1}{2} g \left\{\frac{v_{oy}}{ g}\right\}^2}+ y_o[/tex]

    [tex]y = \frac{v^2_{oy}}{g} \left\{1 - \frac{1}{2}}\right\} + y_o [/tex]

    [tex]y = \frac{v^2_{oy}}{2g} + y_o [/tex]

    Using the componets of your velocities of 34.84 and 36.64 m/s should get the answer.

    Should also note that Y could not exceed [itex] 25\tan{10} +2.1 \leq 6.6 [/itex] just to confirm your numbers are "in the right ball park"
     
  4. Jan 19, 2008 #3
    [​IMG]
     
    Last edited: Jan 19, 2008
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