A particle is thrown up with an initial velocity such that it takes more than one second to reach the top point. What is the distance travelled by the particle during the first second of its decent?

Options

A. `g//2`
B. `g//4`
C. g
D. Information is insufficient to calculate the desired distance.

Explanation

To solve the problem step by step, we will analyze the motion of the particle thrown upwards and then its descent.

### Step 1: Understanding the Motion
When a particle is thrown upwards with an initial velocity, it moves against the gravitational force until it reaches its highest point (the top point). The time taken to reach this top point is given as more than one second.

### Step 2: Time of Ascent and Descent
Let’s denote the time taken to reach the top point as \( t \). Since it takes more than one second to reach the top, we can say \( t > 1 \) seconds. The motion is symmetrical, meaning the time taken to ascend is equal to the time taken to descend. Thus, the time taken to descend from the top point back to the ground is also \( t \) seconds.

### Step 3: Distance Covered in the Last Second of Ascent
To find the distance covered by the particle during the first second of its descent, we need to determine the distance it covered in the last second of its ascent. The last second of ascent occurs between \( t-1 \) seconds and \( t \) seconds.

### Step 4: Using the Equation of Motion
The distance covered during the last second of ascent can be calculated using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S \) is the distance,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \( -g \) for upward motion),
- \( t \) is the time.

For the last second of ascent (from \( t-1 \) to \( t \)):
- The distance covered in the last second can be expressed as:
\[ S = u(t) - u(t-1) + \frac{1}{2}(-g)(1^2) \]
This simplifies to:
\[ S = u - \left(u - g(t-1)\right) - \frac{1}{2}g \]
\[ S = g(t-1) - \frac{1}{2}g \]

### Step 5: Distance During Descent
Due to the symmetry of motion, the distance covered during the first second of descent will be equal to the distance covered during the last second of ascent. Therefore, the distance traveled during the first second of descent is:
\[ S = \frac{1}{2} g \]

### Step 6: Final Calculation
Assuming \( g \approx 9.8 \, \text{m/s}^2 \):
\[ S = \frac{1}{2} \times 9.8 \approx 4.9 \, \text{meters} \]

### Conclusion
The distance traveled by the particle during the first second of its descent is approximately **4.9 meters**.

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