A particlele is projected vertically upwards from the ground with a speed V and a speed v and a second particle is projected at the same instant from a height h directly above the first particel with the same speed v at an angle of projection `theta` with the horizontal in upwards direction. The time when the distance between them is minimum is

Options

A. `(h)/(2v sin theta)`
B. `(h)/(2v cos theta)`
C. `(h)/(sqrt2 v)`
D. `(h)/(2 v)`

Explanation

To find the time when the distance between the two particles is minimum, we can follow these steps:

### Step 1: Understand the Motion of Both Particles
- Particle A is projected vertically upwards from the ground with a speed \( V \).
- Particle B is projected from a height \( h \) directly above particle A with the same speed \( v \) at an angle \( \theta \) with the horizontal.

### Step 2: Write the Equations of Motion
For Particle A (projected vertically):
- The height \( y_A \) at time \( t \) is given by:
\[ y_A = Vt - \frac{1}{2}gt^2 \]

For Particle B (projected at an angle):
- The horizontal and vertical components of the velocity of Particle B are:
- Horizontal: \( v_{Bx} = v \cos(\theta) \)
- Vertical: \( v_{By} = v \sin(\theta) \)

- The height \( y_B \) of Particle B at time \( t \) is given by:
\[ y_B = h + v \sin(\theta) t - \frac{1}{2}gt^2 \]

### Step 3: Find the Relative Position
The vertical distance \( D \) between the two particles at time \( t \) is:
\[ D = y_B - y_A = \left( h + v \sin(\theta) t - \frac{1}{2}gt^2 \right) - \left( Vt - \frac{1}{2}gt^2 \right) \]
This simplifies to:
\[ D = h + (v \sin(\theta) - V)t \]

### Step 4: Find the Minimum Distance
To find the time when the distance \( D \) is minimum, we need to differentiate \( D \) with respect to \( t \) and set the derivative equal to zero:
\[ \frac{dD}{dt} = v \sin(\theta) - V = 0 \]
This gives us:
\[ v \sin(\theta) = V \]

### Step 5: Solve for Time
Now, we can find the time \( t \) when the distance is minimum by substituting back into the equation for \( D \):
\[ D = h + (v \sin(\theta) - V)t \]
Since \( v \sin(\theta) = V \), we have:
\[ D = h \]
The time when the distance is minimum can be expressed as:
\[ t = \frac{h}{2V} \]

### Final Result
Thus, the time when the distance between the two particles is minimum is:
\[ t = \frac{h}{2V} \]