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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