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November 27, 2014

November 27, 2014

Posted by **Emily** on Monday, July 15, 2013 at 9:46pm.

- Math -
**mahesh**, Tuesday, July 16, 2013 at 1:14amv=u-gt or 0=15-9.81t since final velocity is 0 and ball is going against gravity. Thus t=1.53 sec.

s=ut-.5gt^2=15-.5x9.81x(1.53)^2

s=22.95-11.48=121.47 which is the height the ball will go. From there it falls freely for a height of 11.47+1=12.47 ft.

s=0xt+.5x9.81xt^2 or t=(12.47x2/9.81)=1.59 secs.

- Math -
**mahesh**, Tuesday, July 16, 2013 at 1:52amPS: s=11.47 which has appeared as 121.47 at one place in 4th line.

- Math -
**Steve**, Tuesday, July 16, 2013 at 9:55amsince 1 ft = .3 m,

s(t) = .3 + 15t - 4.9t^2

it hits the ground when s=0, so

.3 + 15t - 4.9t^2 = 0

t = 3.08 sec

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