Ball `A` is thrown up vertically with speed `10 m/s`. At the same instant another ball `B` is released from rest at height `h`. At time `t`, the speed of `A` relative to `B` is

Options

A. `10`
B. `10-2 g t`
C. `sqrt(10^(2)-2gh)`
D. `10- g t`

Answer

To solve the problem, we need to find the speed of ball A relative to ball B at a given time \( t \).

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Ball A is thrown upwards with an initial speed \( u_A = 10 \, \text{m/s} \).
- Ball B is released from rest from a height \( h \), so its initial speed \( u_B = 0 \, \text{m/s} \).

2. **Calculate the speed of ball A at time \( t \):**
- The formula for the speed of an object under uniform acceleration is given by:
\[ v = u + at \]
- For ball A, the acceleration \( a \) is \( -g \) (since it is moving upwards against gravity):
\[ v_A = u_A - g t = 10 - g t \, \text{m/s} \]

3. **Calculate the speed of ball B at time \( t \):**
- For ball B, which is falling under the influence of gravity, the initial speed is \( 0 \) and the acceleration is \( g \):
\[ v_B = u_B + g t = 0 + g t = g t \, \text{m/s} \]

4. **Determine the relative speed of ball A with respect to ball B:**
- Since ball A is moving upwards and ball B is moving downwards, their speeds will be added to find the relative speed:
\[ v_{AB} = v_A + v_B \]
- Substituting the expressions we found for \( v_A \) and \( v_B \):
\[ v_{AB} = (10 - g t) + (g t) \]
- Simplifying this gives:
\[ v_{AB} = 10 \, \text{m/s} \]

5. **Conclusion:**
- The relative speed of ball A with respect to ball B at time \( t \) is \( 10 \, \text{m/s} \).

### Final Answer:
The speed of ball A relative to ball B at time \( t \) is \( 10 \, \text{m/s} \).

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