A good quarterback can throw a football at 27 m/s (about 60 mph). How long is the ball in the air?

Assume that

(a) Air resistance is negligible,

(b) g = 9.8 m/s², acceleration due to gravity,

(c) The launch angle is 45°, in order to attain maximum horizontal range.

The horizontal and vertical launch velocities are equal, and each is equal to

(27 m/s) * cos (45°) = 19.09 m/s.

The time to attain maximum height is one half of the time of flight.

Because the vertical velocity is zero at maximum height, half the time of flight, t₁, is given by

19.09 - 9.8t₁ = 0

t₁ = 1.948 s

The time of flight is

2t₁ = 3.896 s

The horizontal distance traveled is

3.896*19.09 = 74.375 m

Answer: The time of flight is 3.9 s (nearest tenth)