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April 21, 2015

April 21, 2015

Posted by **Candice** on Saturday, July 14, 2012 at 4:44am.

Thanks in advance :)

- Math -
**Henry**, Monday, July 16, 2012 at 7:24pmh = ho - 0.5g*t^2 = 0.

4.9 - 4.9*t^2 = 0

-4.9t^2 = -4.9

t^2 = 1

Tf = 1 s. = Fall time or time to

fall to Gnd.

Tr = Tf = 1 s. = Rise time or time to

rise to max. Ht.

h = ho - 0.5g*t^2 = 1.3 m.

4.9 - 4.9*t^2 = 1.3

-4.9t^2 = 1.3 -4.9 = -3.6

t^2 = 0.735

Tf' = 0.857 s. = Fall time or time to fall to 1.3 m.

Dx = Xo * (Tr+Tf').

Dx = 28m/s * 1.857s = 52 m. = Hor. dist.

traveled by the ball.

d = V*T.

V = d/T = (65-52) / 1.857 = 7.0 m/s. =

Speed of fieldsman.

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