B

Step 1: Use the equation of motion to determine the maximum height reached by the projectile. The height (h) can be expressed as:
$$h = \frac{v^2 \sin^2(\theta)}{2g}$$
where

• v = 56 m/s (initial speed),
• g = 9.8 m/s² (acceleration due to gravity),
• h = 40 m (maximum height of the hall).

Step 2: Set the equation equal to the height of the hall:
$$40 = \frac{56^2 \sin^2(\theta)}{2 \cdot 9.8}$$
Step 3: Solve for \( \sin^2(\theta) \):
$$\sin^2(\theta) = \frac{40 \cdot 2 \cdot 9.8}{56^2}$$
Step 4: Calculate the right-hand side:
$$\sin^2(\theta) = \frac{40 \cdot 19.6}{3136} \approx 0.25$$
Step 5: Take the square root:
$$\sin(\theta) = 0.5$$
Step 6: Find the angle:
$$\theta = \sin^{-1}(0.5) = 30^\circ$$
Therefore, the angle at which the ball can be thrown without hitting the ceiling is 30º.