Water drops are falling from a tap at regular time intervals. When the fifth drop is near to fall from tap the first drop is at ground. If the tap is fixed at height H from ground then find out
(i) the height of the second drop from ground
(ii) distance between the second and third drop
(iii) velocity of third drop

Answer

To solve the problem step by step, we will analyze the motion of the water drops falling from the tap.

### Given:
- Drops fall at regular time intervals \( t \).
- When the fifth drop is about to fall, the first drop is on the ground.
- The height of the tap from the ground is \( H \).

### Step 1: Time Analysis
When the fifth drop is about to fall, the first drop has taken a total time of \( 4t \) to reach the ground. Thus, we can conclude:
- The first drop falls for \( 4t \).
- The second drop falls for \( 3t \) (since it fell \( t \) after the first drop).
- The third drop falls for \( 2t \).
- The fourth drop falls for \( t \).
- The fifth drop has just started falling.

### Step 2: Height of the Second Drop from the Ground
To find the height of the second drop from the ground when the first drop is at the ground:
1. The second drop has been falling for \( 3t \).
2. Use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( a = g \) (acceleration due to gravity), and \( t = 3t \).
\[ s_2 = 0 + \frac{1}{2} g (3t)^2 = \frac{1}{2} g \cdot 9t^2 = \frac{9}{2} gt^2 \]
3. The total distance fallen by the first drop (which is at the ground) is:
\[ s_1 = \frac{1}{2} g (4t)^2 = \frac{1}{2} g \cdot 16t^2 = 8gt^2 \]
4. The height of the second drop from the ground is:
\[ H - s_2 = H - \frac{9}{2} gt^2 \]

### Step 3: Distance between the Second and Third Drop
1. The third drop has been falling for \( 2t \):
\[ s_3 = 0 + \frac{1}{2} g (2t)^2 = \frac{1}{2} g \cdot 4t^2 = 2gt^2 \]
2. The distance between the second and third drops is:
\[ \text{Distance} = s_2 - s_3 = \frac{9}{2} gt^2 - 2gt^2 = \frac{9}{2} gt^2 - \frac{4}{2} gt^2 = \frac{5}{2} gt^2 \]

### Step 4: Velocity of the Third Drop
1. The velocity of the third drop after falling for \( 2t \) can be calculated using:
\[ v = u + at \]
where \( u = 0 \) and \( a = g \):
\[ v_3 = 0 + g(2t) = 2gt \]

### Final Answers:
1. Height of the second drop from the ground:
\[ H - \frac{9}{2} gt^2 \]
2. Distance between the second and third drop:
\[ \frac{5}{2} gt^2 \]
3. Velocity of the third drop:
\[ 2gt \]