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March 28, 2017

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A fire hose ejects a stream of water at an angle of 37.9 ° above the horizontal. The water leaves the nozzle with a speed of 25.5 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

i tried 25.5 sin 37.9 but it didn't work

  • physics - ,

    V(ver) = 25.5sin37.9 = 15.66m/s.

    The ver. distance is max when velocity
    is 0:

    Vf = Vo + gt = 0,
    15.66 + (-9.8)t = 0,
    15,66 - 9.8t = 0,
    -9.8d = -15.66,
    t(up) = -15.66 / -9.8 = 1.6s.

    d(ver) = Vo*t + 0.5gt^2,
    d(ver) = 15.66*1.6 + 0.5(-9.8)(1.6)^2,
    d(ver) = 25.06 - 10.39 = 14.67m.

    d(hor) = 14.67 / tan37.9 = 18.84m = Distance from building.

  • physics - ,

    CORRECTION:

    Vo(hor) = 25.5cos37.9 = 20.12m/s.

    Vo(ver) = 25.5sin37.9 = 15.66m/s.

    The ver distance is max when velocity
    is zero:

    Vf = Vo + gt = 0,
    15.66 + (-9.8)t = 0,
    15.66 - 9.8t = 0,
    -9.8t = -15.66,
    t(up) = -15.66 / -9.8 = 1.6s.

    d(hor)=Vo * 2t = 20.12 * 3.2 = 64.3m =
    Distance from building.

  • physics - ,

    i ask my teacher he said it wasn't right

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