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 find the distance traveled by a particle in the first second of its descent after being thrown vertically upward, we can use the equations of motion. Here's a step-by-step solution:

### Step 1: Understand the Motion
When a particle is thrown vertically upward, it will rise until it reaches its maximum height, where its velocity becomes zero. After reaching this point, it will start descending. The distance traveled during the first second of descent is what we need to calculate.

### Step 2: Identify Initial Conditions
At the moment the particle starts its descent:
- The initial velocity (u) at the start of the descent is 0 m/s (as it is at the peak).
- The time (t) for which we want to calculate the distance is 1 second.
- The acceleration due to gravity (g) is approximately 9.8 m/s² (we can use 10 m/s² for simplicity in calculations).

### Step 3: Use the Second Equation of Motion
The second equation of motion states:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \( s \) = distance traveled
- \( u \) = initial velocity
- \( g \) = acceleration due to gravity
- \( t \) = time

### Step 4: Substitute the Values
Since the initial velocity \( u = 0 \) (at the peak), the equation simplifies to:
\[ s = 0 \cdot t + \frac{1}{2}gt^2 \]
\[ s = \frac{1}{2}g(1^2) \]
\[ s = \frac{1}{2}g \]

### Step 5: Calculate the Distance
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ s = \frac{1}{2} \cdot 10 \]
\[ s = 5 \, \text{m} \]

### Conclusion
The distance traveled by the particle in the first second of its descent is **5 meters**.

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