A particle is thrown with any velocity vertically upward, the distance travelled by the particle in first second of its decent is

Options

A. g
B. `(g)/(2)`
C. `(g)/(4)`
D. Cannot be calculated

Answer

To solve the problem of finding the distance traveled by a particle in the first second of its descent after being thrown vertically upward, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
- A particle is thrown vertically upward with an initial velocity \( v \).
- It rises until it reaches the highest point, where its velocity becomes \( 0 \).
- After reaching the highest point, the particle begins its descent.

2. **Identify the Initial Conditions for Descent**:
- At the highest point, the initial velocity \( u \) for the descent is \( 0 \) m/s (since it momentarily stops before falling down).
- The acceleration \( g \) (due to gravity) acts downward and has a value of approximately \( 9.81 \, \text{m/s}^2 \).

3. **Use the Kinematic Equation**:
- We use the equation of motion for the distance traveled under uniform acceleration:
\[ s = ut + \frac{1}{2}gt^2 \]
- Here, \( s \) is the distance traveled, \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time.

4. **Substitute the Known Values**:
- Since \( u = 0 \) (initial velocity at the highest point) and \( t = 1 \) second (the time of descent we are interested in), we can substitute these values into the equation:
\[ s = 0 \cdot 1 + \frac{1}{2} \cdot g \cdot (1)^2 \]
- This simplifies to:
\[ s = \frac{1}{2}g \]

5. **Calculate the Distance**:
- Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ s = \frac{1}{2} \cdot 9.81 \approx 4.905 \, \text{m} \]

6. **Conclusion**:
- The distance traveled by the particle in the first second of its descent is approximately \( 4.905 \, \text{m} \).

### Final Answer:
The distance traveled by the particle in the first second of its descent is approximately \( 4.905 \, \text{m} \).