C

Step 1: Analyze the initial velocity of the ball.
The boy throws the ball with a muzzle velocity (V) of 20 m/s at an angle of 37º. We can resolve this velocity into horizontal and vertical components:
Horizontal component, \( V_x = V \cdot \cos(37º) = 20 \cdot \frac{4}{5} = 16 \text{ m/s} \)
Vertical component, \( V_y = V \cdot \sin(37º) = 20 \cdot \frac{3}{5} = 12 \text{ m/s} \)

Step 2: Determine the time of flight.
For vertical motion, we can use the formula for vertical displacement. Since the boy catches the ball at the same height, the vertical displacement is zero:
\( t = \frac{2V_y}{g} = \frac{2 \cdot 12}{10} = 2.4 \text{ seconds} \)

Step 3: Analyze horizontal motion.
While the ball is in the air, it moves with the horizontal velocity \( V_x = 16 \text{ m/s} \), and we need to find how far the train has moved in this time under constant acceleration. Let's assume the train's acceleration is \( a \) and the initial speed is 10 m/s:
Distance moved by the train, \( d_{train} = 10t + \frac{1}{2} a t^2 = 10(2.4) + \frac{1}{2} a (2.4^2) \)

The horizontal motion of the ball relative to the train should allow the boy to catch it back, which means by the time the boy catches the ball, its horizontal displacement concerning the train should be zero. Thus, the displacement must equal the displacement of the ball:
\( d_{ball} = ,V_x \cdot t = 16 \cdot 2.4 \)

The equation for distance equality can be set to find the speed of the train \( V_{train} \) when the ball is caught:

Setting the equations equal to each other, \( 10(2.4) + \frac{1}{2} a (2.4^2) = 16(2.4) \)

Step 4: Solve for acceleration and final velocity of the train.
By solving the above equation, we get the necessary values. Therefore, substituting the values will yield the final speed calculation and determination of options provided will be (A, B, C, D). After calculating and checking options, the correct answer can be concluded as C.