A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the changes in kinetic energy (KE) and potential energy (PE) of a raindrop falling from a height \( h \) above the ground, particularly as it falls through a height of \( \frac{3}{4}h \) and reaches near terminal velocity.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- At the height \( h \) (point C), the raindrop has maximum potential energy and zero kinetic energy.
- The potential energy (PE) at this point is given by:
\[ PE = mgh \]
- The kinetic energy (KE) at this point is:
\[ KE = 0 \]

2. **Analyze the Fall to Height \( \frac{3}{4}h \):**
- As the raindrop falls from height \( h \) to \( \frac{3}{4}h \) (point B), it loses potential energy and gains kinetic energy.
- The potential energy at height \( \frac{3}{4}h \) is:
\[ PE = mg\left(\frac{3}{4}h\right) = \frac{3}{4}mgh \]
- The change in potential energy from point C to point B is:
\[ \Delta PE = PE_{initial} - PE_{final} = mgh - \frac{3}{4}mgh = \frac{1}{4}mgh \]
- The kinetic energy at this point (point B) can be calculated using the work-energy principle. The work done on the raindrop is equal to the change in kinetic energy:
\[ KE_{B} = \Delta PE = \frac{1}{4}mgh \]

3. **Analyze the Fall to the Ground:**
- As the raindrop continues to fall from height \( \frac{3}{4}h \) to the ground (point A), it reaches near terminal velocity.
- At the ground (point A), the potential energy is zero:
\[ PE = 0 \]
- The kinetic energy at this point (point A) is maximum and can be calculated as:
\[ KE_{A} = KE_{B} + \Delta PE = \frac{1}{4}mgh + \frac{3}{4}mgh = mgh \]

4. **Summarize Energy Changes:**
- From point C to point B:
- PE decreases from \( mgh \) to \( \frac{3}{4}mgh \)
- KE increases from \( 0 \) to \( \frac{1}{4}mgh \)
- From point B to point A:
- PE decreases from \( \frac{3}{4}mgh \) to \( 0 \)
- KE increases from \( \frac{1}{4}mgh \) to \( mgh \)

5. **Conclusion:**
- The correct diagram will show a decrease in potential energy and an increase in kinetic energy. Initially, potential energy is high and kinetic energy is low, and as the drop falls, potential energy decreases while kinetic energy increases until it reaches the ground.