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March 29, 2017

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I am trying to solve a problem using Newton's equation, which is (-1/2)(9.8)t^2+ vot+ h. The problem says that a car rolls of a 120 foot high cliff. I know how to right the equation for this problem which would be (-1/2)(9.8 m/s) t^2 + vot + 37m. I am trying to find out how high the car will befter 2 seconds.

  • math - ,

    well, when it starts off the cliff it may have horizontal velocity but has no vertical velocity so Vo = 0
    (1/2) * 9.8 = 4.9 so we might write that as
    z = -4.9 t^2 + 0 + 37
    so after two seconds we have:
    z = -4.9 * 4 + 37
    z = 37-19.6
    or 17.4 meters above ground
    ==================================
    by the way, the vertical velocity would be:
    V = Vo - 4.9 t
    V = 0 - 9.8
    or 9.8 meters per second down
    The horizontal velocity would be whatever horizontal velocity it started with because there is no acceleration in the horizontal plane, only vertical.

  • whoops, correction - ,

    *

    by the way, the vertical velocity would be:
    V = Vo - 9.8 t
    V = 0 - 19.6
    or 19.6 meters per second down
    The horizontal velocity would be whatever horizontal velocity it started with because there is no acceleration in the horizontal plane, only vertical.

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  • whoops, correction - ,

    Thank you.

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