A particle is projected vertically upward with velocity `u` from a point `A`, when it returns to the point of projection .

Options

A. Its average speed is `u//2`.
B. Its average velocity is zero.
C. Its displacement is zero.
D. Its average speed is `u`.

Explanation

To solve the problem of a particle projected vertically upward with an initial velocity `u` from point A and returning to the same point, we will follow these steps:

### Step 1: Understand the Motion
When the particle is projected upwards, it will rise to a maximum height and then fall back down to the original point A. The total distance traveled by the particle is the distance it goes up plus the distance it comes back down.

### Step 2: Calculate Displacement
Displacement is defined as the shortest distance between the initial and final positions. Since the particle returns to the original point A, the displacement is:
\[ \text{Displacement} = 0 \]

### Step 3: Calculate Total Distance
The total distance traveled by the particle consists of two segments:
1. The distance traveled upwards (let's denote this distance as `s`).
2. The distance traveled downwards, which is also `s`.

Thus, the total distance \( D \) is:
\[ D = s + s = 2s \]

### Step 4: Calculate the Distance Upwards (s)
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \) (the final velocity at the maximum height),
- \( u \) is the initial velocity,
- \( a = -g \) (acceleration due to gravity, negative because it acts downwards).

Substituting these values, we get:
\[ 0 = u^2 - 2gs \]
Rearranging gives:
\[ s = \frac{u^2}{2g} \]

### Step 5: Calculate Total Distance
Now substituting \( s \) into the total distance:
\[ D = 2s = 2 \left(\frac{u^2}{2g}\right) = \frac{u^2}{g} \]

### Step 6: Calculate Total Time Taken
Using the first equation of motion:
\[ v = u + at \]
For the upward journey:
\[ 0 = u - gt \]
Solving for \( t \) gives:
\[ t = \frac{u}{g} \]
This is the time taken to reach the maximum height. The time taken to come back down is the same, so the total time \( T \) is:
\[ T = 2t = 2 \left(\frac{u}{g}\right) = \frac{2u}{g} \]

### Step 7: Calculate Average Velocity
Average velocity \( V_{avg} \) is defined as:
\[ V_{avg} = \frac{\text{Displacement}}{\text{Total Time}} \]
Substituting the values:
\[ V_{avg} = \frac{0}{\frac{2u}{g}} = 0 \]

### Step 8: Calculate Average Speed
Average speed \( S_{avg} \) is defined as:
\[ S_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Substituting the values:
\[ S_{avg} = \frac{\frac{u^2}{g}}{\frac{2u}{g}} = \frac{u^2}{g} \cdot \frac{g}{2u} = \frac{u}{2} \]

### Final Results
- Displacement = 0
- Average Velocity = 0
- Total Distance = \( \frac{u^2}{g} \)
- Average Speed = \( \frac{u}{2} \)