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Originally Posted by JugglingReferee
Camron,
I see your formula: 4•t², but how did derive that formula? My physics days are more behind me than I thought, or it truly is 7am on Sunday morning...
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My calculation is based on the fact that at the peak, the velocity of the ball is 0 and that, for these distances/speeds, it takes essentially half the time to go up and half the time to go do down.
I use the following:
vy0 = 0...velocity at the peak of the throw
g = 32.2 (gravity at sea level)
t = 1, 1.5, and 2 (half of the observed time estimates, T)
So, the equation simplifes to 16.1 * t² (fall time) or 4.025 * T² (total time),
Someone could pull out a stopwatch and time it exactly if they wish and calculate it to the exact inch if they want.