Physics
posted by Allen .
During a baseball game, a batter hits a popup to a fielder 83 m away.
The acceleration of gravity is 9.8 m/s^2.
If the ball remains in the air for 6.5 s, how high does it rise?
Answer in units of m

)A batter hits a popup to a fielder 83 m away. If the ball remains in the air for 6.5 s, how high does it rise?
Rise time = fall time = 3.25sec.
Vf = 0 = Vo  9.8(3.25) or Vo = 31.85m/s
d = Vo^2(sin(2µ))/g
31.85^2(sin(2µ))/9.8 or µ = 26.65º
Then, height h = Vo^2(sin^(x)/2g or
h = 31.85^2(sin(26.65))/2(9.8) = 10.41m 
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