posted by ra .
During a baseball game, a batter hits a popup to a ﬁelder 64 m away.
The acceleration of gravity is 9.8 m/s
If the ball remains in the air for 6.7 s, how
high does it rise?
Answer in units of m
Dx = Xo*6.7 = 64.
Xo = 9.55 m/s = Hor. component of initial velocity.
Tr = Tf = 6.7/2 = 3.35 s. = Rise and Fall time.
Y = Yo + gt = 0.
Yo - 9.8*3.35 = 0
Yo = 32.83 m/s. = Ver. component of
h = Yo*t + 0.5g*t^2.
h = 32.83*3.35 -4.9*(3.35)^2 = 55 m.