A

Given the horizontal velocity of the cart, \( v_1 = 2 \, \text{m/s} \), and the vertical velocity of the rain drops, \( v_2 = 6 \, \text{m/s} \).

To find the angle \( \alpha \) with respect to the horizontal, we use the concept of relative velocities. The pipe will align with the effective velocity of the raindrops relative to the moving cart.

The effective velocity of the raindrops can be represented by a resultant vector from the horizontal and vertical components which can be computed as follows:

• The horizontal component (velocity of the cart): \( v_{horizontal} = v_1 = 2 \, \text{m/s} \)
• The vertical component (velocity of the rain): \( v_{vertical} = v_2 = 6 \, \text{m/s} \)

The angle \( \alpha \) can be represented using the tangent function:
$$ \tan(\alpha) = \frac{v_{vertical}}{v_{horizontal}} = \frac{6}{2} = 3 $$

Therefore, to find the angle \( \alpha \):
$$ \alpha = \tan^{-1}(3) \approx 71.57^\circ $$

Thus, the angle at which the pipe should be placed is approximately \( 71.57^\circ \). Since the question asks for the angle to the horizontal, we accept this value as the final answer.