A body is dropped from the top of the tower and falls freely.
The velocity of the body after `n` seconds is proportional to .

Options

A. `n^(2)`
B. `n`
C. `2n-1`
D. `2n^(2) -1`

Explanation

To solve the problem, we need to determine how the velocity of a body falling freely from a tower changes with time. Here’s a step-by-step solution:

### Step 1: Understand the scenario
A body is dropped from the top of a tower, meaning it starts from rest. The initial velocity (u) of the body is 0 m/s. The body falls under the influence of gravity, which provides a constant acceleration (g).

### Step 2: Identify the relevant equations
We will use the first equation of motion, which relates initial velocity, final velocity, acceleration, and time:
\[ v = u + at \]
where:
- \( v \) = final velocity after time \( t \)
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 3: Substitute the known values
Since the body is dropped:
- \( u = 0 \) m/s (initial velocity)
- \( a = g \) (acceleration due to gravity)
- \( t = n \) seconds (time after which we want to find the velocity)

Substituting these values into the equation:
\[ v = 0 + g \cdot n \]
\[ v = g \cdot n \]

### Step 4: Analyze the relationship
From the equation \( v = g \cdot n \), we can see that the velocity \( v \) is directly proportional to \( n \) (since \( g \) is a constant).

### Conclusion
Thus, the velocity of the body after \( n \) seconds is proportional to \( n \).

### Final Answer
The velocity of the body after \( n \) seconds is proportional to \( n \).

---