At a particular height, the velocity of an ascending body is `vecu`. The velocity at the same height while the body falls freely is

Options

A. `2vecu`
B. `-vecu`
C. `vecu`
D. `-2vecu`

Explanation

To solve the problem, we need to analyze the situation of an object ascending and another object descending at the same height. Let's break it down step by step.

### Step 1: Understand the scenario
We have two objects:
1. An ascending body with a velocity `\(\vec{u}\)` at a height `h`.
2. A descending body that is falling freely under the influence of gravity.

### Step 2: Identify the forces and energy
The ascending body has kinetic energy due to its velocity and potential energy due to its height. The descending body, while falling freely, has potential energy at the same height and gains kinetic energy as it falls.

### Step 3: Apply the law of conservation of energy
According to the law of conservation of energy, the total mechanical energy (kinetic + potential) of both bodies at the same height should be equal.

1. For the ascending body:
- Kinetic Energy (KE) = \(\frac{1}{2} m u^2\)
- Potential Energy (PE) = \(mgh\)

Total energy at height `h` = \(\frac{1}{2} m u^2 + mgh\)

2. For the descending body:
- At the same height, the potential energy is also \(mgh\).
- Let the velocity of the descending body at height `h` be `\(\vec{v}\)`.

Total energy at height `h` = \(\frac{1}{2} m v^2 + mgh\)

### Step 4: Set the total energies equal
Since the total mechanical energy must be conserved, we can set the total energies equal to each other:

\[ \frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2 + mgh \]

### Step 5: Simplify the equation
We can cancel out the potential energy terms \(mgh\) from both sides:

\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 \]

### Step 6: Solve for the velocity of the descending body
Now, we can simplify further by canceling out the mass \(m\) and the factor of \(\frac{1}{2}\):

\[ u^2 = v^2 \]

Taking the square root of both sides gives us:

\[ |u| = |v| \]

### Step 7: Determine the direction of the velocity
Since the ascending body has a velocity `\(\vec{u}\)` in the upward direction, the descending body will have the same magnitude of velocity but in the opposite direction. Therefore, we can express the velocity of the descending body as:

\[ \vec{v} = -\vec{u} \]

### Final Answer
The velocity of the descending body at the same height is:

\[ \vec{v} = -\vec{u} \]