A body is projected vertically upward with speed `10 m//s` and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take `g = 10 m//s^(2)`}

Options

A. Zero
B. `10 m//s^(2)`
C. `5m//s^(2)`
D. `20m//s^(2)`

Answer

To solve the problem, we need to determine the magnitude of the acceleration of the first body (projected upward) with respect to the second body (projected downward).

### Step-by-Step Solution:

1. **Identify the Acceleration of Each Body**:
- The first body (let's call it Body A) is projected upward with an initial speed of \(10 \, \text{m/s}\). The only force acting on it after projection is gravity, which accelerates it downward at \(g = 10 \, \text{m/s}^2\).
- The second body (let's call it Body B) is projected downward from the top of the tower with the same initial speed of \(10 \, \text{m/s}\). It also experiences the same gravitational acceleration downward at \(g = 10 \, \text{m/s}^2\).

2. **Determine the Acceleration of Each Body**:
- The acceleration of Body A (upward) is \(a_A = -g = -10 \, \text{m/s}^2\) (negative because it is upward).
- The acceleration of Body B (downward) is \(a_B = g = 10 \, \text{m/s}^2\).

3. **Calculate the Relative Acceleration**:
- The relative acceleration of Body A with respect to Body B is given by the formula:
\[ a_{AB} = a_A - a_B \]
- Substituting the values:
\[ a_{AB} = (-10 \, \text{m/s}^2) - (10 \, \text{m/s}^2) = -20 \, \text{m/s}^2 \]

4. **Determine the Magnitude of the Relative Acceleration**:
- The magnitude of the relative acceleration is:
\[ |a_{AB}| = 20 \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the acceleration of the first body with respect to the second body is \(20 \, \text{m/s}^2\).

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