A balloon rises up with constant net acceleration of `10 m//s^(2)`. After 2 s a particle drops from the balloon. After further 2 s match the following : `("Take" g=10 m//s^(2))`
`{:(,"Column I",,,"Column II"),((A),"Height of perticle from ground",,(p),"Zero"),((B),"Speed of particle",,(q),10 SI "units"),((C),"Displacement of Particle",,(r),40 SI "units"),((D),"Acceleration of particle",,(s),20 SI "units"):}`

Explanation

To solve the problem step by step, we need to analyze the motion of the balloon and the particle that drops from it.

### Step 1: Analyze the motion of the balloon
The balloon rises with a constant net acceleration of \(10 \, \text{m/s}^2\). We need to find the velocity of the balloon after 2 seconds.

Using the first equation of motion:
\[ V = U + at \]
Where:
- \(U = 0 \, \text{m/s}\) (initial velocity of the balloon)
- \(a = 10 \, \text{m/s}^2\) (acceleration of the balloon)
- \(t = 2 \, \text{s}\)

Substituting the values:
\[ V = 0 + (10 \times 2) = 20 \, \text{m/s} \]

### Step 2: Calculate the height of the balloon after 2 seconds
Using the second equation of motion to find the displacement \(S_1\) of the balloon:
\[ S = Ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ S_1 = 0 \times 2 + \frac{1}{2} \times 10 \times (2^2) = 0 + \frac{1}{2} \times 10 \times 4 = 20 \, \text{m} \]

### Step 3: Analyze the motion of the particle after it drops
When the particle drops from the balloon after 2 seconds, it has an initial velocity equal to the velocity of the balloon at that moment, which is \(20 \, \text{m/s}\) upward.

### Step 4: Determine the motion of the particle in the next 2 seconds
The particle will experience downward acceleration due to gravity (\(g = 10 \, \text{m/s}^2\)). We need to find the final velocity \(V\) of the particle after 2 seconds of free fall.

Using the first equation of motion:
\[ V = U + at \]
Where:
- \(U = 20 \, \text{m/s}\) (initial velocity of the particle)
- \(a = -10 \, \text{m/s}^2\) (acceleration due to gravity, negative because it is downward)
- \(t = 2 \, \text{s}\)

Substituting the values:
\[ V = 20 + (-10 \times 2) = 20 - 20 = 0 \, \text{m/s} \]

### Step 5: Calculate the displacement of the particle during the 2 seconds
Using the second equation of motion to find the displacement \(S_2\) of the particle:
\[ S = Ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ S_2 = 20 \times 2 + \frac{1}{2} \times (-10) \times (2^2) = 40 - 20 = 20 \, \text{m} \]

### Step 6: Total height of the particle from the ground
The total height of the particle from the ground is the height of the balloon plus the displacement of the particle:
\[ \text{Height} = S_1 + S_2 = 20 + 20 = 40 \, \text{m} \]

### Step 7: Final results
1. Height of particle from ground: \(40 \, \text{m}\)
2. Speed of particle after 4 seconds: \(0 \, \text{m/s}\)
3. Displacement of particle: \(0 \, \text{m}\) (since it returns to the same height it was dropped)
4. Acceleration of particle: \(10 \, \text{m/s}^2\) downward (due to gravity)

### Summary of Results
- Height of particle from ground: \(40 \, \text{m}\) (matches with option (r))
- Speed of particle: \(0 \, \text{m/s}\) (matches with option (p))
- Displacement of particle: \(0 \, \text{m}\) (matches with option (q))
- Acceleration of particle: \(10 \, \text{m/s}^2\) (matches with option (s))