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September 2, 2014

September 2, 2014

Posted by **Chris** on Tuesday, May 14, 2013 at 1:27am.

- math -
**Henry**, Wednesday, May 15, 2013 at 5:30pmV = 103Ft/s[34o] + 22Ft./s[0o].

Xo = 103*cos34 + 22*cos0 = 107.4 Ft/s.

Yo = 103*sin34 + 22*sin(00 = 57.6 Ft/s.

tan A = Y/X = 57.6/107.4 = 0.53628

A = 28.2o.

V = Xo/cosA = 107.4/cos28.2 = 121.6 Ft/s.[28.2o]

Y^2 = Yo^2 + 2g*h

hmax = ho + (Y^2-Yo^2)/2g

hmax = 3 + (0-(57.6)^2)/-64=54.84 Ft

above gnd.

Tr = (Y-Yo)/g = (0-57.6)/-32 = 1.8 s. =

Rise time.

h = Yo*t + 0.5g*t^2 = 54.84-10

0 + 16t^2 = 44.84

t^2 = 2.8

Tf = 1.67 s. = Time to fall to 10 Ft above gnd.

Range = Xo * (Tr+Tf)=107.4*(1.8+1.67) =

373 Ft. Yes, it clears the fence.

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