Two particles are released from the same height at an interval of `1s`. How long aftger 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. Let's break it down step by step.

### Step 1: Define the variables
Let:
- \( t \) = time in seconds after the first particle is released.
- The second particle is released 1 second later, so its time of fall is \( t - 1 \).

### Step 2: Use the second law of motion
The distance fallen by an object under free fall is given by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since both particles are released from rest, the initial velocity \( u = 0 \). The acceleration \( a = g = 10 \, \text{m/s}^2 \) (downward).

### Step 3: Calculate the distance for the first particle
For the first particle (released at \( t = 0 \)):
\[ s_1 = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
Substituting \( g = 10 \):
\[ s_1 = \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2 \]

### Step 4: Calculate the distance for the second particle
For the second particle (released at \( t = 1 \)):
\[ s_2 = 0 \cdot (t - 1) + \frac{1}{2} g (t - 1)^2 = \frac{1}{2} g (t - 1)^2 \]
Substituting \( g = 10 \):
\[ s_2 = \frac{1}{2} \cdot 10 \cdot (t - 1)^2 = 5(t - 1)^2 \]

### Step 5: Set up the equation for the distance apart
According to the problem, the distance between the two particles when they are 10 meters apart is given by:
\[ s_1 - s_2 = 10 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ 5t^2 - 5(t - 1)^2 = 10 \]

### Step 6: Simplify the equation
Factor out the common term:
\[ 5(t^2 - (t - 1)^2) = 10 \]
Dividing both sides by 5:
\[ t^2 - (t - 1)^2 = 2 \]

### Step 7: Expand and simplify
Expanding \( (t - 1)^2 \):
\[ t^2 - (t^2 - 2t + 1) = 2 \]
This simplifies to:
\[ t^2 - t^2 + 2t - 1 = 2 \]
\[ 2t - 1 = 2 \]

### Step 8: Solve for \( t \)
Adding 1 to both sides:
\[ 2t = 3 \]
Dividing by 2:
\[ t = 1.5 \, \text{s} \]

### Final Answer
The two particles will be 10 meters apart after **1.5 seconds** from the moment the first particle begins to fall.

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