Posted by **Anonymous** on Monday, September 17, 2012 at 3:22pm.

A golf ball is dropped from rest from a height of 9.60 m. It hits the pavement, then bounces back up, rising just 5.30 m before falling back down again. A boy then catches the ball when it is 1.11 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

- physics -
**Steve**, Monday, September 17, 2012 at 3:55pm
to drop 9.6m takes √(9.6/4.9)=1.40 seconds

to rise 5.3m takes √(5.3/4.9)=1.04 seconds

to fall 4.19m takes √(4.19/4.9)=0.92 seconds

so, the ball is in the air 3.36 seconds

- physics -
**Anonymous**, Thursday, October 13, 2016 at 5:42pm
This is wrong, u cant use t=sqrt of (2d/g) because that is only for when Vo is 0 m/s, when the ball hits the ground the velocity is not 0m/s, therefore when it comes back up the Vo is not 0m/s, the true answer is about 2.32s

- physics -
**Anonymous**, Thursday, October 13, 2016 at 5:42pm
This is wrong, u cant use t=sqrt of (2d/g) because that is only for when Vo is 0 m/s, when the ball hits the ground the velocity is not 0m/s, therefore when it comes back up the Vo is not 0m/s, the true answer is about 2.32s

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