A particle is thrown vertically upwards with a velocity `v`. It returns to the ground in time T. which of the following graphs correctly represents the motion ?

Options

A.
B.
C.
D.

Explanation

To solve the problem of a particle thrown vertically upwards with an initial velocity \( v \) and returning to the ground in time \( T \), we need to analyze the motion of the particle step by step.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a particle is thrown upwards, it moves against the force of gravity. The initial velocity is \( v \), and it will decelerate due to gravitational acceleration \( g \) until it reaches its maximum height.

2. **Time to Reach Maximum Height**:
- The time taken to reach the maximum height (where the velocity becomes zero) is given by:
\[ t_{\text{up}} = \frac{v}{g} \]

3. **Total Time of Flight**:
- The total time of flight \( T \) is the time taken to go up and come back down. Since the time taken to go up is equal to the time taken to come down:
\[ T = 2 \cdot t_{\text{up}} = 2 \cdot \frac{v}{g} \]

4. **Velocity vs. Time Graph**:
- The velocity of the particle changes over time. Initially, it starts with a positive velocity \( v \) and decreases to 0 at \( t = \frac{T}{2} \) (the maximum height). After that, it starts increasing in the negative direction (downwards) until it reaches the ground at \( t = T \).
- The graph will start at \( v \), decrease linearly to 0 at \( t = \frac{T}{2} \), and then linearly increase in the negative direction until it reaches \( -v \) at \( t = T \).

5. **Choosing the Correct Graph**:
- The correct graph should show a linear decrease from \( v \) to 0 and then a linear increase from 0 to \( -v \). This corresponds to a straight line that slopes downwards and then slopes upwards in the negative direction.

### Conclusion:
The correct graph that represents the motion of the particle is option A, which shows the velocity decreasing to 0 and then increasing in the negative direction.

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