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.

Answer 1

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.