If `v` be the instantaneous velocity of the body dropped from the top of a tower, when it is located at height `h`, then which of the following remains constant?

Options

A. `gh+v^(2)`
B. `gh+(v^(2))/(2)`
C. `gh-(v^(2))/(2)`
D. `gh-v^(2)`

Explanation

To solve the problem, we need to analyze the situation of a body being dropped from a height \( h \) and determine what remains constant during its fall.

### Step-by-Step Solution:

1. **Understanding the System**:
- A body is dropped from a height \( h \).
- As it falls, it experiences gravitational acceleration, which affects its velocity.

2. **Identifying Energy Types**:
- The total mechanical energy of the system consists of two forms: kinetic energy (KE) and potential energy (PE).
- Kinetic Energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Potential Energy (PE) at height \( h \) is given by:
\[ PE = mgh \]

3. **Total Mechanical Energy**:
- The total mechanical energy \( E \) of the body at height \( h \) can be expressed as:
\[ E = KE + PE = \frac{1}{2} mv^2 + mgh \]

4. **Conservation of Mechanical Energy**:
- According to the principle of conservation of mechanical energy, in the absence of air resistance and other non-conservative forces, the total mechanical energy remains constant throughout the motion.
- Therefore, we can write:
\[ \frac{1}{2} mv^2 + mgh = \text{constant} \]

5. **Analyzing the Equation**:
- Since \( m \) (mass of the body) is constant, we can factor it out:
\[ \frac{1}{2} v^2 + gh = \text{constant} \]
- This indicates that the sum of the kinetic energy per unit mass and the potential energy per unit mass remains constant.

6. **Conclusion**:
- The quantity that remains constant as the body falls is the total mechanical energy, which can be expressed as:
\[ \frac{1}{2} v^2 + gh = \text{constant} \]
- Therefore, the correct answer is that the total mechanical energy remains constant.