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Physics

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A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 19.0 m/s. The cliff is h = 50.0 m above a flat horizontal beach.

With what speed and angle of impact does the stone land?

  • Physics - ,

    find the vertical speed at impact:

    vf^2=2*g*h
    You know the vertical velocity, and the horizontal velocity.

    From that, you get the angle, and the magnitude.

  • Physics - ,

    I was able to find out the angle which is 57.83degree that is correct.
    For the speed, i used sqrt(2*9.81*50)=31.32, is this correct because the hw site doesn't accept it. What am I missing?
    Thanks.

  • Physics - ,

    The sqrt(2*9.81*50)= 31.32 (m/s) that you calculated is the vertical velocity compnent. There is also a horizontal component (19.0 m/s)

    You need to combine them with the Pythagorean equatiuon to get the speed.

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