V = 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.
A major leaugue baseball player hits a ball 3 feet above the ground with a velocity of 103 feet per secon in the direction of a 10 foot wall that is 300 feet from home plate. If the hit is at an angle of elevation of 34 degrees and there is wind blowing 22 feet per second in the SAME direction horizontally,, determine if the ball clears the fence.
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