BYU - Washington

[JugglingReferee]

Originally Posted by Camron Rust

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:

v_y0 = 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.

I recall now - thanks! We were taught that the imperial value for the acceleration due to gravity is 32.2 ft/s². We used 9.80665 m/s².