A

Step 1: Analyzing the dropped particle (A)
The first particle is dropped from a height of 10 m. It is subjected to gravitational acceleration only. The time taken to fall can be calculated using the equation of motion:
$h = \frac{1}{2} g t^2$
Where:
- $h$ is the height (10 m)
- $g$ is the acceleration due to gravity (approximately 9.81 m/s²)
- $t$ is the time taken to fall.
Substituting the values, we have:
$$10 = \frac{1}{2} \times 9.81 \times t^2$$
Solving for $t$, we rearrange the equation:
$$t^2 = \frac{10 \times 2}{9.81} \approx 2.03$$
Taking the square root:
$$t \approx \sqrt{2.03} \approx 1.42 \, \text{seconds}$$

Step 2: Analyzing the horizontally projected particle (B)
The second particle is projected horizontally at a speed of 5 m/s. Although it is moving horizontally, it is also affected by gravity as it falls from the same height of 10 m. The time taken to fall to the ground is the same as for particle A, since vertical motion is independent of horizontal motion.
Using the same equation as above, we find that the time taken $t$ for this particle to reach the ground is also approximately 1.42 seconds.

Conclusion:
Since both particles take the same time of approximately 1.42 seconds to reach the ground, we conclude that both particles reach the ground simultaneously.
Therefore, the correct option is A.