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-   -   Acceleration due to gravity (https://www.askmehelpdesk.com/showthread.php?t=322912)

  • Feb 27, 2009, 04:51 PM
    v1nc3nzo87
    Acceleration due to gravity
    State why the acceleration due to gravity can be found using the slope of the graph.
  • Feb 27, 2009, 05:02 PM
    ROLCAM

    Extract from Wikipedia:-

    The strength of the gravitational field is numerically equal to the acceleration of objects under its influence, and its value at the Earth's surface, denoted g, is approximately expressed below as the standard average.

    g = 9.8 m/s2 = 32.2 ft/s2

    This means that, ignoring air resistance, an object falling freely near the earth's surface increases its velocity with 9.8 m/s (32.2 ft/s or 22 mph) for each second of its descent. Thus, an object starting from rest will attain a velocity of 9.8 m/s (32.2 ft/s) after one second, 19.6 m/s (64.4 ft/s) after two seconds, and so on, adding 9.8 m/s (32.2 ft/s) to each resulting velocity. Also, again ignoring air resistance, any and all objects, when dropped from the same height, will hit the ground at the same time.
    The above can be easily portrayed in a graphical way.
  • Feb 28, 2009, 08:19 AM
    sarnian
    Hello v1nc3nzo87

    Just repeating parts of Wikipedia text here still does not explain why the acceleration due to gravity can be found using the slope of the graph.
    In the previous post we only see a line vertical up and down again.
    Now if we add some horizontal force component, you get a nice graph, a slope of a graph.
    Remember (figure an airplane or a bullet) : the horizontal force may change over time, but the vertical (G) force remains.

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