A body A is thrown vertically upwards with initial velocity `v_(1)`. Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t if the bodies begin to move simultaneously.

Options

A. `x= h -v_(1)t`
B. `x = (h-v_(1))t`
C. `x=h - (v_(1))/(t)`
D. `x= (h)/(t) - v_(1)`

Explanation

To solve the problem, we need to analyze the motion of both bodies A and B, and then find the distance \( x \) between them as a function of time \( t \).

### Step-by-Step Solution:

1. **Identify the motions of the two bodies**:
- Body A is thrown upwards with an initial velocity \( v_1 \).
- Body B is dropped from a height \( h \) with an initial velocity of 0.

2. **Write the equation of motion for Body A**:
- The displacement \( h_1 \) of Body A after time \( t \) can be given by the equation:
\[ h_1 = v_1 t - \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity, and we take upward direction as positive.

3. **Write the equation of motion for Body B**:
- The displacement \( h_2 \) of Body B after time \( t \) can be given by:
\[ h_2 = \frac{1}{2} g t^2 \]
- Since Body B is falling downwards, we consider its displacement as positive in the downward direction.

4. **Relate the distances**:
- The total height \( h \) can be expressed as:
\[ h = h_1 + x + h_2 \]
- Rearranging gives:
\[ x = h - h_1 - h_2 \]

5. **Substitute the expressions for \( h_1 \) and \( h_2 \)**:
- Substitute \( h_1 \) and \( h_2 \) into the equation for \( x \):
\[ x = h - \left(v_1 t - \frac{1}{2} g t^2\right) - \frac{1}{2} g t^2 \]
- Simplifying this gives:
\[ x = h - v_1 t + \frac{1}{2} g t^2 - \frac{1}{2} g t^2 \]
- The \( \frac{1}{2} g t^2 \) terms cancel out:
\[ x = h - v_1 t \]

6. **Final expression**:
- The distance \( x \) between the two bodies as a function of time \( t \) is:
\[ x = h - v_1 t \]

### Conclusion:
The distance \( x \) between the two bodies depends linearly on time \( t \) and is given by the equation:
\[ x = h - v_1 t \]