The relevant equations we can use in this problem are:
d = v0 t + 0.5 a t^2
v^2 = v0^2 + 2 a d
a. how high does it go?
We are to find for the distance d. The formula is:
v^2 = v0^2 + 2 a d
where v is final velocity = 0 at the peak, v0 = 25 m/s, a is negative gravity since opposite to direction = - 9.8 m/s^2, d is height achieved = ?
0 = (25)^2 + 2 (-9.8) d
d = 31.89 m
Therefore the maximum height is about 31.89 meters
b. how long is it in the air
We use the formula:
d = v0 t + 0.5 a t^2
in going up, a = -g = -9.8 m/s^2, find for time t:
31.89 = 25 t + 0.5 (- 9.8) t^2
-4.9 t^2 + 25 t = 31.89
t^2 – 5.1 t = 6.51
Completing the square:
(t – 2.55)^2 = 6.51 + 6.5025
t – 2.55 = ± 3.61
t = -1.06, 6.16
Since time cannot be negative, therefore:
t = 6.16 seconds
However this is not the total time yet, because this only accounts for the time going up.
The time going up and going down are equal therefore:
t(total) = 2 t
t(total) = 12.32 seconds
Therefore the foul ball spent 12.32 seconds in the air.