Posted by **kayla** on Thursday, March 24, 2011 at 8:13pm.

from 4 feet above a swimming pool, susan throws a ball upward witha velocity of 32 feet per second. how long does it take the ball to reach the water?

- math -
**Henry**, Sunday, March 27, 2011 at 3:35am
(Vf)^2 = Vo^2 + 2gd = 0,

(32)^2 - 2*32d = 0,

1024 - 64d = 0,

-64d = -1024,

d(up) = 16ft.

Vf = Vo + gt = 0,

32 - 32t = 0,

-32t = -32,

t(up) = 1s.

d(down) = 4 + 16 = 20ft.

d = Vo + 0.5gt^2 = 20ft.

d = 0 + 0.5*32t^2 = 20,

16t^2 = 20,

t^2 = 1.25,

t(down) = 1.12s.

t(up) + t(down) = 1 + 1.12 = 2.12s =

time for ball to reach water.

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