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Posted by on Saturday, July 14, 2012 at 4:38am.

Hi, could you please help with this question :)

Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 m high and 30 m away.

Thanks in advance

  • Math - , Monday, July 16, 2012 at 4:33pm

    h1 = 15 m.
    h2 = 26.25 - 15 = 11.25 m. = Ver. dist.
    from firing point to top of wall.

    Y = h2 = 11.25 m.
    X = 30 m.
    tanA = Y/X = 11.25 / 30 = 0.375
    A = 20.6 Deg. = Direction of particle.

    Range = Vo^2*sin(2A)/g = 30 m.
    Vo^2*sin(41.2) / 9.8 = 30
    Divide both sides by sin(41.2):
    Vo^2 / 9.8 = 45.5
    Vo^2 = 446.3
    Vo = 21.1 m/s @ 20.6 Deg. = Velocity
    and direction.

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