Two bodies begin a free fall from rest from the same height 2 seconds apart. How long after the first body begins to fall, the two bodies will be 40 m apart? `("Take g = 10ms"^(-2))`
1 s
2 s
3 s
4 s

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Answer

To solve the problem, we need to determine how long after the first body begins to fall the two bodies will be 40 meters apart. Let's break it down step by step.

### Step 1: Understand the motion of the two bodies
- The first body is dropped at \( t = 0 \) seconds.
- The second body is dropped at \( t = 2 \) seconds.

### Step 2: Use the equations of motion
For free fall, the distance \( s \) traveled by an object from rest under gravity \( g \) is given by:
\[ s = \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \).

### Step 3: Calculate the distance traveled by the first body
Let \( t \) be the time in seconds after the first body is dropped. The distance \( s_1 \) traveled by the first body after \( t \) seconds is:
\[ s_1 = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times t^2 = 5t^2 \]

### Step 4: Calculate the distance traveled by the second body
The second body is dropped 2 seconds later, so it falls for \( t - 2 \) seconds. The distance \( s_2 \) traveled by the second body is:
\[ s_2 = \frac{1}{2} g (t - 2)^2 = \frac{1}{2} \times 10 \times (t - 2)^2 = 5(t - 2)^2 \]

### Step 5: Set up the equation for the distance between the two bodies
According to the problem, the distance between the two bodies is 40 meters:
\[ s_1 - s_2 = 40 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ 5t^2 - 5(t - 2)^2 = 40 \]

### Step 6: Simplify the equation
Expanding \( (t - 2)^2 \):
\[ (t - 2)^2 = t^2 - 4t + 4 \]
Thus,
\[ s_2 = 5(t^2 - 4t + 4) = 5t^2 - 20t + 20 \]
Now substituting back into the equation:
\[ 5t^2 - (5t^2 - 20t + 20) = 40 \]
This simplifies to:
\[ 20t - 20 = 40 \]
Adding 20 to both sides:
\[ 20t = 60 \]
Dividing by 20:
\[ t = 3 \]

### Conclusion
The two bodies will be 40 meters apart after **3 seconds** of the first body beginning to fall.

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