The maximum horizontal distance can be achieved when the ball is thrown at an angle of 45 degrees with the horizontal. This is because at this angle, the vertical component of the velocity is equal to the horizontal component, resulting in the maximum range.

Since the ball cannot hit the ceiling, we need to find the maximum angle that allows the ball to stay below the height of the ceiling (40 m).

Using the equation for the vertical motion of the ball:
h = ut + (1/2)gt^2

Where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

Let's assume the ball reaches its maximum height at time t/2 (half the total time of flight).

At this time, the vertical displacement is equal to the height of the ceiling (40 m):
40 = (u * t/2) + (1/2)(-9.8)(t/2)^2

Simplifying the equation:
80 = u * t - 4.9 * t^2

We know the initial vertical velocity is given by u * sin(theta), where theta is the launch angle.

Solving for u:
u = 56 / sin(theta)

Substituting this into the equation:
80 = (56 / sin(theta)) * t - 4.9 * t^2

To find the maximum angle, we want to find the angle that allows the ball to stay below the height of the ceiling for the longest time.

Let's differentiate the equation with respect to time and set it equal to zero to find the maximum time:
0 = (56 / sin(theta)) - 9.8 * t

Solving for t:
t = (56 / (9.8 * sin(theta)))

Substituting this into the equation:
80 = (56 / sin(theta)) * (56 / (9.8 * sin(theta))) - 4.9 * (56 / (9.8 * sin(theta)))^2

Simplifying the equation:
80 = 32 / sin^2(theta) - (32 / sin^2(theta))^2

Let's simplify further by substituting sin^2(theta) = 1 - cos^2(theta):
80 = 32 / (1 - cos^2(theta)) - (32 / (1 - cos^2(theta)))^2

Let's substitute x = cos^2(theta):
80 = 32 / (1 - x) - (32 / (1 - x))^2

Multiplying both sides by (1 - x)^2:
80(1 - x)^2 = 32(1 - x) - 32^2

Expanding and simplifying:
80 - 160x + 80x^2 = 32 - 32x - 1024

Rearranging the equation:
80x^2 + 128x - 1120 = 0

Solving this quadratic equation, we find two solutions for x: x = -2 and x = 7/5.

Since cos^2(theta) cannot be negative, we discard the x = -2 solution.

Therefore, cos^2(theta) = 7/5.

Solving for cos(theta):
cos(theta) = sqrt(7/5)

Taking the inverse cosine to find theta:
theta = cos^(-1)(sqrt(7/5))

Using