A football is kicked at ground level with a speed of 18.8 m/s at an angle of 41.0 ∘ to the horizontal.?

The time is dependent on the football’s initial vertical velocity.
Initial vertical velocity = 18.8 * sin 41

This is approximately 12.3 m/s. During one half of the total time, the vertical velocity decreases from this velocity to 0 m/s at the rate of 9.8 m/s each second. Let’s use the following equation to determine the time for this to happen.

vf = vi – g * t
0 = 18.8 * sin 41 – 9.8 * t
t = 18.8 * sin 41 ÷ 9.8

This is approximately 1.26 seconds. The total time is twice this number.
Total time = 37.6 * sin 41 ÷ 9.8
This is approximately 2.52 seconds.

Take g = 9.81 m/s²
Take all upward quantities to be positive, and downward quantities to be negative.

Consider the vertical component of the
u = 18.8 sin41.0° m/s
a = -9.81 m/s²
s = 0 m

s = ut + (1/2)at²
0 = (18.8 sin41.0°)*t + (1/2)*(-9.81)*t²
(1/2)*9.81*t² = (18.8 sin41.0°)*t
t ≠ 0, then (1/2)*9.81*t = (18.8 sin41.0°)
Time taken, t = (18.8 sin41.0°)*2/9.81 s = 2.51 s

Initial vertical V component = (sin 41 x 18.8) = 12.33m/sec.
Time in air = 2(v/g) = 2(12.33/9.8) = 2.52 secs.

2.5 s

Vertical motion:

y = 18.8sin(41)t - 1/2gt^2

The ball hits the ground when y = 0

i. e 18.8sin(41)t - 1/2gt^2 = 0

37.6sin(41) = gt

t = 37.6sin(41)/9.8

= 2.517124438

= 2.52 s

fly time i twice the time t' to achieve the maximum height
t' is given by
0 = v0y - g t' ==> t' = v0y / g
then
t = 2 t' = 2 v0y / g
t = 2*18.8*sin41 / 9.8 = 2.52 s