Two particles are released from the same height at an interval of `1s`. How long after the first particle begins to fall will the two particles be `10m` apart? (`g=10m//s^(2)`)

Options

A. `1.5s`
B. `2s`
C. `1.25s`
D. `2.5s`

Answer

To solve the problem, we need to determine how long after the first particle begins to fall the two particles will be 10 meters apart.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Particle 1 is released at time \( t = 0 \).
- Particle 2 is released at time \( t = 1 \) second.
- We need to find the time \( t \) after the release of Particle 1 when the distance between the two particles is 10 meters.

2. **Using the Equations of Motion**:
- The distance fallen by an object under gravity (with initial velocity \( u = 0 \)) is given by the equation:
\[ s = ut + \frac{1}{2} g t^2 \]
- For both particles, since they are released from rest, the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

3. **Calculating the Distance Fallen by Each Particle**:
- For Particle 1 (released at \( t = 0 \)):
- At time \( t \), the distance fallen is:
\[ H_1 = \frac{1}{2} g t^2 \]
- For Particle 2 (released at \( t = 1 \)):
- At time \( t \), the time since its release is \( t - 1 \). Thus, the distance fallen is:
\[ H_2 = \frac{1}{2} g (t - 1)^2 \]

4. **Setting Up the Equation**:
- According to the problem, the distance between the two particles is 10 meters:
\[ H_1 - H_2 = 10 \]
- Substituting the expressions for \( H_1 \) and \( H_2 \):
\[ \frac{1}{2} g t^2 - \frac{1}{2} g (t - 1)^2 = 10 \]

5. **Simplifying the Equation**:
- Factor out \( \frac{1}{2} g \):
\[ \frac{1}{2} g \left( t^2 - (t - 1)^2 \right) = 10 \]
- Expanding \( (t - 1)^2 \):
\[ (t - 1)^2 = t^2 - 2t + 1 \]
- Therefore:
\[ t^2 - (t^2 - 2t + 1) = 2t - 1 \]
- Substituting back:
\[ \frac{1}{2} g (2t - 1) = 10 \]

6. **Substituting \( g = 10 \, \text{m/s}^2 \)**:
- Plugging in the value of \( g \):
\[ \frac{1}{2} \times 10 (2t - 1) = 10 \]
- Simplifying:
\[ 5(2t - 1) = 10 \]
- Dividing both sides by 5:
\[ 2t - 1 = 2 \]
- Solving for \( t \):
\[ 2t = 3 \quad \Rightarrow \quad t = \frac{3}{2} = 1.5 \, \text{seconds} \]

### Final Answer:
The two particles will be 10 meters apart after **1.5 seconds** from the release of the first particle.

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