Water drops fall from a tap on to the floor 5.0 m below at regular intervals of time. The first drop strikes the floor when the fifth drops beings to fall. The height at which the third drop will be from ground at the instant when the first drop strikes the ground is (take`g=10 m^(-2)`)

Options

A. `1.25m`
B. `2.15m`
C. `2.75 m`
D. `3.75 m`

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration due to gravity.

### Step 1: Understand the problem
We have a tap from which water drops fall at regular intervals, and the height from the tap to the ground is 5.0 m. We need to find the height of the third drop from the ground when the first drop hits the ground.

### Step 2: Determine the time taken for the first drop to fall
Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s = 5 \, \text{m} \) (distance fallen),
- \( u = 0 \, \text{m/s} \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t \) is the time taken to fall.

Substituting the values:
\[ 5 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 5 = 5t^2 \]
Dividing both sides by 5:
\[ t^2 = 1 \]
Taking the square root:
\[ t = 1 \, \text{s} \]

### Step 3: Determine the time interval between drops
Since the first drop strikes the ground when the fifth drop begins to fall, the time taken for the first drop to fall is equal to the time taken for the fourth drops to fall before the fifth drop starts falling. Thus:
\[ T = 4t \]
Where \( T \) is the time for the first drop to fall (1 s), and \( t \) is the time interval between successive drops.

Substituting \( T \):
\[ 1 = 4t \]
Solving for \( t \):
\[ t = \frac{1}{4} \, \text{s} = 0.25 \, \text{s} \]

### Step 4: Find the time at which the third drop falls
The third drop starts falling after the first drop has fallen for \( 2t \):
\[ t_3 = 2t = 2 \times 0.25 = 0.5 \, \text{s} \]

### Step 5: Calculate the displacement of the third drop after 0.5 s
Now, we need to find how far the third drop has fallen after 0.5 seconds using the same equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \( u = 0 \),
- \( g = 10 \, \text{m/s}^2 \),
- \( t = 0.5 \, \text{s} \).

Substituting the values:
\[ s = 0 \cdot 0.5 + \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]
This simplifies to:
\[ s = \frac{1}{2} \cdot 10 \cdot 0.25 \]
\[ s = \frac{10}{2} \cdot 0.25 = 1.25 \, \text{m} \]

### Step 6: Calculate the height of the third drop from the ground
The height of the third drop above the ground when the first drop strikes the ground is:
\[ h' = H - s \]
Where:
- \( H = 5 \, \text{m} \) (initial height),
- \( s = 1.25 \, \text{m} \) (displacement of the third drop).

Substituting the values:
\[ h' = 5 - 1.25 = 3.75 \, \text{m} \]

### Final Answer
The height at which the third drop will be from the ground when the first drop strikes the ground is **3.75 m**.

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