Answer:
3.82 s
Explanation:
The football follows a projectile path, where the x-component of the velocity will remain constant, and the y-component of the velocity will vary because of gravity.
The time elapsed before the ball hits the ground is 2 times the time for the ball to reach its maximum height, which is also the time necessary for the vertical velocity of the ball to go from its original value to 0:
[tex]V_y = V*sin(58.5) = 22m/s*sin(58.5) = 18.76 m/s\\\\a = \frac{v_f - v_o}{t} \\t_{up} = \frac{v_f - v_o}{g} = \frac{0m/s - 18.76m/s}{-9.81 m/s^2} = 1.912 s\\\\t_{air} = 2*t{up} = 3.82s[/tex]
