Posted by Sahil Patel on Wednesday, September 7, 2011 at 6:42pm.
A baseball player releases a baseball at the height of 7 feet above the ground with an initial vertical velocity of 54 feet per second. How long will it take the ball to reach the ground?

algebra  Henry, Saturday, September 10, 2011 at 3:06pm
Vf^2 = Vo^2 + 2gd,
d = (Vf^2  Vo^2) / 2g,
d = 7 + (0  (54)^2) / 64 = 52.56 Ft.
= Distance above ground.
Vf = Vo + gt,
t(up) = (Vf  Vo) / g,
t(up) = (0  54) / 32 = 1.69s.
d = Vo*t + 0.5gt^2 = 52.56 Ft.
0 + 0.5*32t = 52.56,
16t = 52.56,
t(down) = 3.29s.
T = t(up) + t(down) = 1.69 + 3.29 = 4.98s = Time to reach ground.

algebra  Henry, Saturday, September 10, 2011 at 3:19pm
Correction:
0 + 0.5*32t^2 = 52.56,
16t^2 = 52.56,
t^2 = 3.29,
t(down) = 1.81s.
T = t(up) + t(down) = 1.69 + 1.81 = 3.50s. = Time to reach ground.
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