Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant. `(g = 10 ms^-2)`

Options

A. `1.25m`
B. `2.50m`
C. `3.75m`
D. `4.00m`

Explanation

To solve the problem, we need to determine the height of the second drop from the ground at the instant the first drop touches the ground. Let's break down the steps:

### Step 1: Determine the time taken for the first drop to reach the ground

The first drop falls from a height of 5 meters. We can use the second equation of motion to find the time taken for it to reach the ground:
\[ S = ut + \frac{1}{2} a t^2 \]

Here:
- \( S = 5 \) meters (distance fallen)
- \( u = 0 \) (initial velocity)
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) is the time taken

Substitute the known values:
\[ 5 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 5 = 5 t^2 \]
\[ t^2 = 1 \]
\[ t = 1 \, \text{second} \]

### Step 2: Determine the time intervals between drops

Since the drops fall at regular intervals and the third drop is leaving the tap when the first drop touches the ground, the time interval between each drop is:
\[ t_0 = \frac{t}{2} = \frac{1}{2} \, \text{second} \]

### Step 3: Determine the height of the second drop

The second drop has been falling for \( t_0 = 0.5 \) seconds. We need to find the distance it has fallen in this time using the same equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]

Here:
- \( u = 0 \) (initial velocity)
- \( a = g = 10 \, \text{m/s}^2 \)
- \( t = t_0 = 0.5 \) seconds

Substitute the known values:
\[ S = 0 \cdot 0.5 + \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]
\[ S = \frac{1}{2} \cdot 10 \cdot 0.25 \]
\[ S = 1.25 \, \text{meters} \]

### Step 4: Calculate the height of the second drop from the ground

The height of the second drop from the ground is the initial height minus the distance it has fallen:
\[ H = 5 - 1.25 \]
\[ H = 3.75 \, \text{meters} \]

### Final Answer
The height of the second drop from the ground at the instant the first drop touches the ground is \( 3.75 \) meters.

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