Analysis:
To solve this problem, we need to understand the concept of free fall and the equations of motion. When an object is dropped, it falls freely due to the force of gravity. The distance it travels vertically can be determined using the equations of motion.

Key Points:
- Drops falling from a tap at regular intervals
- Tap is 5 m above the ground
- Third drop leaves the tap when the first drop touches the ground
- Find the distance of the second drop from the ground when the first drop touches the ground

Solution:
Let's assume that the time taken by each drop to reach the ground is the same, and we'll call it 't' seconds. We'll also assume that the initial velocity of each drop is zero.

Using the equations of motion:
1. The distance traveled by an object in free fall can be calculated using the equation:
s = ut + (1/2)gt^2
where s is the distance traveled, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

2. Since the initial velocity is zero (u = 0), the equation simplifies to:
s = (1/2)gt^2

Calculating the distance of the first drop:
The first drop falls freely from a height of 5 m, so the distance traveled by the first drop is:
s1 = (1/2)gt^2

Calculating the distance of the second drop:
The second drop falls freely from a height of 5 m - s1 (distance remaining after the first drop falls). So, the distance traveled by the second drop is:
s2 = (1/2)g(t-t1)^2
where t1 is the time taken by the first drop to reach the ground.

Calculating the time taken by the first drop:
The time taken by the first drop to reach the ground is the same as the time taken by the third drop to leave the tap. Therefore, t1 = t.

Substituting the values:
s2 = (1/2)g(t-t)^2
= (1/2)g(0)^2
= 0

Conclusion:
The distance of the second drop from the ground when the first drop touches the ground is 0 meters. This means that the second drop is at the same level as the ground at that instant.

Final Answer:
The second drop is at ground level when the first drop touches the ground.