Physics
posted by Logan .
A ball is thrown directly upward with an initial velocity of 14 m/s. If the ball is released from an initial height of 2.8 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.

Vf = Vo^2 + 2gd,
d(up) = (Vf^2  Vo^2) / 2g,
d(up) = (0  (14)^2) / 19.6 = 10m.
D = 2.8 + 10 = 12.8m above ground.
Vf = Vo + gt,
t(up) = (Vf  Vo) / g,
t(up) = (0  14) / 9.8 = 1.43s.
D = Vo*t + 0.5gt^2 = 12.8,
0 + 0.5*9.8t^2 = 12.8,
4.9t^2 = 12.8,
t^2 = 2.61,
t(down) = 1.62s.
T = 1.43 + 1.62 = 3.05s. in air.
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