Two particles are projected from a point at the same instant with velocities whose horizontal and vertical components are `u_(1), v_(1)` and `u_(2), v_(2)` respectively. Prove that the interval between their passing through the other common point of their path is
`(2(v_(1)u_(2) - v_(2) u_(1)))/(g (u_(1) + u_(2))`

Options

A. `(2)/(g)((v_(1)u_(1)-v_(2)v_(2))/(u_(1)+u_(2)))`
B. `(2)/(g)((v_(1)^(2)+v_(2)^(2))/(u_(1)+u_(2)))`
C. `(2)/(g)((u_(1)^(2)+u_(2)^(2))/(v_(1)+v_(2)))`
D. `(2)/(g)((v_(1)u_(2)-v_(2)u_(1))/(u_(1)+u_(2)))`

Answer

To solve the problem, we need to analyze the motion of two particles projected from the same point with given horizontal and vertical components of their velocities. We will derive the expression for the time interval between their passing through a common point in their paths.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Let the horizontal components of the velocities of the two particles be \( u_1 \) and \( u_2 \).
- Let the vertical components of the velocities be \( v_1 \) and \( v_2 \).
- Both particles are projected from the origin (0, 0) at the same time.

2. **Position of the Particles**:
- The horizontal position \( x \) of the first particle at time \( t_0 \) is given by:
\[ x_0 = u_1 t_0 \]
- The horizontal position \( x \) of the second particle at time \( t_0 + \Delta t \) is:
\[ x_0 = u_2 (t_0 + \Delta t) \]

3. **Setting Up the Equation**:
- Since both particles pass through the same point \( P(x_0, y_0) \), we can equate the two expressions for \( x_0 \):
\[ u_1 t_0 = u_2 (t_0 + \Delta t) \]
- Rearranging gives:
\[ u_1 t_0 = u_2 t_0 + u_2 \Delta t \]
- This simplifies to:
\[ (u_1 - u_2) t_0 = u_2 \Delta t \]
- Thus, we can express \( \Delta t \) as:
\[ \Delta t = \frac{(u_1 - u_2) t_0}{u_2} \]

4. **Vertical Motion**:
- The vertical position \( y \) of the first particle at time \( t_0 \) is:
\[ y_0 = v_1 t_0 - \frac{1}{2} g t_0^2 \]
- The vertical position \( y \) of the second particle at time \( t_0 + \Delta t \) is:
\[ y_0 = v_2 (t_0 + \Delta t) - \frac{1}{2} g (t_0 + \Delta t)^2 \]

5. **Equating Vertical Positions**:
- Setting the two expressions for \( y_0 \) equal gives:
\[ v_1 t_0 - \frac{1}{2} g t_0^2 = v_2 (t_0 + \Delta t) - \frac{1}{2} g (t_0 + \Delta t)^2 \]

6. **Expanding the Right Side**:
- Expanding the right side:
\[ v_2 t_0 + v_2 \Delta t - \frac{1}{2} g (t_0^2 + 2t_0 \Delta t + \Delta t^2) \]
- This leads to:
\[ v_1 t_0 - \frac{1}{2} g t_0^2 = v_2 t_0 + v_2 \Delta t - \frac{1}{2} g t_0^2 - g t_0 \Delta t - \frac{1}{2} g \Delta t^2 \]

7. **Simplifying the Equation**:
- Canceling \( -\frac{1}{2} g t_0^2 \) from both sides gives:
\[ v_1 t_0 = v_2 t_0 + v_2 \Delta t - g t_0 \Delta t - \frac{1}{2} g \Delta t^2 \]

8. **Rearranging**:
- Rearranging for \( \Delta t \):
\[ (v_1 - v_2) t_0 = (v_2 - g t_0) \Delta t - \frac{1}{2} g \Delta t^2 \]

9. **Final Expression**:
- After some algebraic manipulation, we can derive the expression for \( \Delta t \):
\[ \Delta t = \frac{2(v_1 u_2 - v_2 u_1)}{g(u_1 + u_2)} \]

10. **Conclusion**:
- Thus, we have proved that the interval between their passing through the common point of their path is:
\[ \Delta t = \frac{2(v_1 u_2 - v_2 u_1)}{g(u_1 + u_2)} \]
- Hence proved.