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On 23 Oct 2001, at 15:39, Larry Cartwright wrote:
Even a quick look at the equations of motion, without any solving,
confirms that the thrown ball travels farther than the falling ball.
The distances traveled by the two balls is y1 = 1/2*a*t^2 (falling) and
y2 = 1/2*a*t^2 + v0*t (thrown). The second distance y2 is obviously
greater than the first distance y1 (for any positive v0 and relatively
small value of t).
Substitute v0 = at and we get the distance travelled by the body
going up as 3/2*a*t^2. This shows that the distance travelled by it
is three times that of the one coming down.
regards,
Sarma.
Hope this helps,
Larry
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Larry Cartwright <exit60@cablespeed.com>
Retired (June 2001) Physics Teacher
Charlotte MI 48813 USA
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"Information is not knowledge,
knowledge is not wisdom,
and wisdom is not foresight.
Each grows out of the other
and we need them all."
(Arthur C. Clarke)
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