C

Step 1: Use the formula for the maximum height of a projectile:
$$ H = \frac{v^2 \sin^2(\theta)}{2g} $$
where
- $H$ is the maximum height
- $v$ is the initial velocity (40 m/s in this case)
- $g$ is the acceleration due to gravity (approximately 9.81 m/s²)

Step 2: For the first ball (height $H_1$):
$$ H_1 = \frac{(40)² \sin^2(\theta)}{2 \times 9.81} $$
Step 3: For the second ball (height $H_2$):
$$ H_2 = \frac{(40)² \sin^2(90° - \theta)}{2 \times 9.81} $$
Since $\sin(90° - \theta) = \cos(\theta)$, we have:
$$ H_2 = \frac{(40)² \cos^2(\theta)}{2 \times 9.81} $$
Step 4: Given that $H_2 = H_1 + 50$:
$$ \frac{(40)² \cos^2(\theta)}{2 \times 9.81} = \frac{(40)² \sin^2(\theta)}{2 \times 9.81} + 50 $$
Step 5: Simplifying, we find:
$$ 50 = H_2 - H_1 $$ leads to
$$ 50 = \frac{(40)²}{2 \times 9.81}(\cos^2(\theta) - \sin^2(\theta)) $$
After finding the required angles and calculating heights, the values arrive as:
$H_1 = 15 m$, $H_2 = 65 m$
Therefore, option C is correct.