A soccer ball is kicked from the playing field at a 45° angle. If the ball is in the air for 3 s, what is the maximum height achieved?

Hello!
Denote the angle as alpha, the initial speed as V and the given time as T.
I suppose we ignore air resistance. Then the only force acting on the ball is the gravity force, it is directed downwards and gives the acceleration g = 9.8 m/s^2 to the ball.
The vertical component of the velocity uniformly decreases with time t from V sin(alpha) with the speed g, so it is equal to V sin(alpha) - g t. The height itself is equal to H(t) = V sin(alpha) t - (g t^2)/2. At the time T the velocity is zero, i.e. V sin(alpha)T =(g T^2)/2, or V sin(alpha) = (g T)/2.
The maximum height is reached when the vertical speed becomes zero, i.e. when V sin(alpha) = g t. From the above we know that this time is T/2.
Finally, the maximum height is
H(T/2) =Vsin(alpha) T/2 - (g T^2)/8 =(g T^2)/4 -(g T^2)/8 =(g T^2)/8.
Numerically it is (9.8*9)/8 approx 11 (m). This is the answer. Note that it doesn't depend on alpha.