A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. How long is she in the air before returning to Earth? (g = 9.81 m/s2)

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  1. Vo = 12m/s[20o]
    Xo = 12*cos20 = 11.28 m/s.
    Yo = 12*sin20 = 4.10 m/s.

    Y = Yo + g*Tr = 0 @ max. ht.
    4.10 - 9.8Tr = 0
    -9.8Tr = -4.10
    Tr = 0.419 s. = Rise time.

    Tf = Tr = 0.419 s.

    T = Tr + Tf = 0.419 + 0.419 = 0.838 s.=
    Time in air.

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  2. 13m

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