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November 29, 2014

November 29, 2014

Posted by **ruby** on Monday, July 21, 2008 at 7:21pm.

- How could you calculate how high the ball goes and how long the ball is in the air before it comes to the thrower's hand?

- physics -
**drwls**, Monday, July 21, 2008 at 7:39pmIt travels upwards until the vertical velocity component becomes zero. An easier way to calculate it it to require that the initial kinetic energy become completely potential energy there.

(1/2) M Vo^2 = M g H

where H is the maximum height above the thrower's hand.

H = Vo^2/(2 g) = 11.5 m

The time in the air is twice the time it takes for the velocity upward to become zero. It spends an equal length of time coming back down. Therefore:

T = 2 * (Vo/g) = 3.06 s

- physics -
**ruby**, Monday, July 21, 2008 at 8:42pmthanx so much! im finally starting to get this!

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