A foul ball is hit straight up into the air with a speed of about 25 m/s.? how high does it go? (b) how long is it in the air?

To solve this problem, we make use of the equations of motion:

vf^2 = v0^2 - 2 g h

vf = v0 - g t

h = v0 t - 0.5 g t^2

A. Finding for max height:

Max height occurs when vf = 0, therefore:

0^2 = 25^2 - 2 (9.8) h

h = 31.89 m

B. Finding for t:

t = t (up) + t (down)

vf = v0 - g t (up)

0 = 25 - 9.8 t (up)

t (up) = 2.55 s

h = v0 t (down) - 0.5 g t (down) ^2

31.89 = 0.5 (9.8) t (down) ^2

t (down) ^2 = 6.51

t (down) = 2.55 s

so,

t = 5.1 s