This content has been marked as final.
Show 2 replies

acceleration
superres Sep 24, 2003 12:00 AM (in response to fiftyfivedisabled)On 9/24/2003 7:04:36 PM, fiftyfive wrote:
>What is the acceleration of a
>rock at the top of its
>trajectory when thrown
>straight upward? Explain
>whether or not the answer is
>zero by using the equation
>a=F/m as a guide to your
>thinking.
And how far have you gotten on this?
Do you think that gravity shuts off when your rock reaches the top of its trajectory?
TTFN,
Eden 
acceleration
ptc1295782 Sep 25, 2003 12:00 AM (in response to fiftyfivedisabled)On 9/24/2003 7:04:36 PM, fiftyfive wrote:
>What is the acceleration of a
>rock at the top of its
>trajectory when thrown
>straight upward? Explain
>whether or not the answer is
>zero by using the equation
>a=F/m as a guide to your
>thinking.

y(t) is a continuous function of time. Acceleration in the second derivative of position with respect to time. Since acceleration is related to the curvature of y(t), acceleration is non zero at the top of the trajectory. Since the curvature is concave downward, the acceleration component in the y direction is negative.
Since the y component of the force, F, is negative and constant for small vertical displacements near the surface of Earth, and since the mass is constant at terrestrial speeds, therefore the ratio F/m is a negative constant and is equal to the y component of the acceleration. This is true no matter where the object is in the trajectory, as long as it is in free fall.

Physics: Common Sense made Obscure by Mathematics Don Sparlin