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

Options

A. `t^0`
B. `t^1`
C. `t^2`
D. `1/t`

Answer

To solve the problem of the relative speed of two balls—one dropped from a height and the other thrown upwards—we can follow these steps:

### Step 1: Identify the initial conditions of both balls.
- **Ball A** (dropped from a height of 45 m):
- Initial velocity (u_A) = 0 m/s (since it is dropped)
- Acceleration (a_A) = g = 9.81 m/s² (downward)

- **Ball B** (thrown upwards):
- Initial velocity (u_B) = 40 m/s (upward)
- Acceleration (a_B) = -g = -9.81 m/s² (since it is moving against gravity)

### Step 2: Write the equations of motion for both balls.
- For Ball A (dropped):
\[ h_A(t) = h_0 - \frac{1}{2} g t^2 \]
where \( h_0 = 45 \) m is the initial height.

- For Ball B (thrown upwards):
\[ h_B(t) = u_B t - \frac{1}{2} g t^2 \]
where \( u_B = 40 \) m/s.

### Step 3: Determine the relative speed of the balls.
The relative speed of two objects is given by the difference in their velocities. The velocity of each ball can be derived from their equations of motion.

- Velocity of Ball A (v_A):
\[ v_A(t) = 0 + g t = g t \]

- Velocity of Ball B (v_B):
\[ v_B(t) = u_B - g t = 40 - g t \]

Now, the relative velocity (v_rel) of Ball A with respect to Ball B is:
\[ v_{rel}(t) = v_A(t) - v_B(t) = (g t) - (40 - g t) = 2g t - 40 \]

### Step 4: Substitute the value of g.
Using \( g = 9.81 \) m/s²:
\[ v_{rel}(t) = 2(9.81)t - 40 = 19.62t - 40 \]

### Conclusion:
The relative speed of the balls varies linearly with time and is given by:
\[ v_{rel}(t) = 19.62t - 40 \text{ m/s} \]