A ball ` A` is dropped from a building of height ` 45 m`. Simultaneously another ball ` B` is thrown up with a speed ` 40 m//s`. Calculate the relative speed of the balls as a function of time.

Options

A. 0
B. `10 m s^(-1)`
C. `25 m s^(-1)`
D. `50 m s^(-1)`

Explanation

To solve the problem of finding the relative speed of two balls, A and B, as a function of time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Ball A is dropped from a height of 45 m, so its initial velocity \( u_A = 0 \) m/s.
- Ball B is thrown upwards with an initial velocity \( u_B = 40 \) m/s.

2. **Determine the Acceleration**:
- Both balls experience gravitational acceleration \( g = 9.81 \) m/s² (downwards).

3. **Write the Velocity Equations**:
- For ball A (falling down):
\[ v_A(t) = u_A + (-g)t = 0 - gt = -gt \]
- For ball B (moving upwards):
\[ v_B(t) = u_B + (-g)t = 40 - gt \]

4. **Calculate the Relative Velocity**:
- The relative velocity of A with respect to B is given by:
\[ v_{AB}(t) = v_A(t) - v_B(t) \]
- Substitute the expressions for \( v_A(t) \) and \( v_B(t) \):
\[ v_{AB}(t) = (-gt) - (40 - gt) \]
- Simplifying this:
\[ v_{AB}(t) = -gt - 40 + gt = -40 \text{ m/s} \]

5. **Interpret the Result**:
- The relative speed \( v_{AB}(t) = -40 \) m/s indicates that ball A is moving downwards relative to ball B at a constant speed of 40 m/s.

### Final Answer:
The relative speed of the balls as a function of time is \( -40 \) m/s, which means that ball A is falling downwards at a speed of 40 m/s relative to ball B.