To solve the problem step by step, we will analyze the motion of the balloon and the particle dropped from it.
### Step 1: Calculate the height of the balloon after 2 seconds.
The balloon rises with a constant acceleration of \(10 \, \text{m/s}^2\). We can use the formula for displacement:
\[
h = ut + \frac{1}{2} a t^2
\]
where:
- \(u = 0 \, \text{m/s}\) (initial velocity of the balloon),
- \(a = 10 \, \text{m/s}^2\) (acceleration),
- \(t = 2 \, \text{s}\) (time).
Substituting the values:
\[
h = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 = 0 + \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{m}
\]
### Step 2: Calculate the speed of the balloon after 2 seconds.
We can use the formula for final velocity:
\[
v = u + at
\]
where:
- \(u = 0 \, \text{m/s}\),
- \(a = 10 \, \text{m/s}^2\),
- \(t = 2 \, \text{s}\).
Substituting the values:
\[
v = 0 + 10 \cdot 2 = 20 \, \text{m/s}
\]
### Step 3: Analyze the motion of the particle after it is dropped.
When the particle is dropped from the balloon, it has an initial velocity of \(20 \, \text{m/s}\) (upward) and will experience a downward acceleration due to gravity (\(g = 10 \, \text{m/s}^2\)).
### Step 4: Calculate the height of the particle after 2 seconds.
We can use the formula for displacement of the particle:
\[
h = ut - \frac{1}{2} g t^2
\]
where:
- \(u = 20 \, \text{m/s}\) (initial velocity of the particle),
- \(g = 10 \, \text{m/s}^2\),
- \(t = 2 \, \text{s}\).
Substituting the values:
\[
h = 20 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2)^2 = 40 - \frac{1}{2} \cdot 10 \cdot 4 = 40 - 20 = 20 \, \text{m}
\]
### Step 5: Calculate the height of the particle from the ground.
Since the balloon was at a height of \(20 \, \text{m}\) after 2 seconds, and the particle dropped from it, the total height of the particle from the ground is:
\[
\text{Height from ground} = \text{Height of balloon} + \text{Height of particle} = 20 + 20 = 40 \, \text{m}
\]
### Step 6: Calculate the speed of the particle after 2 seconds.
Using the formula for final velocity:
\[
v = u - gt
\]
where:
- \(u = 20 \, \text{m/s}\),
- \(g = 10 \, \text{m/s}^2\),
- \(t = 2 \, \text{s}\).
Substituting the values:
\[
v = 20 - 10 \cdot 2 = 20 - 20 = 0 \, \text{m/s}
\]
### Step 7: Calculate the acceleration of the particle.
The particle is in free fall after being dropped, so it experiences an acceleration equal to \(g\) (downward):
\[
\text{Acceleration} = g = 10 \, \text{m/s}^2
\]
### Summary of Results:
- Height of particle from ground: \(40 \, \text{m}\)
- Speed of particle: \(0 \, \text{m/s}\)
- Displacement of particle: \(20 \, \text{m}\) (from the point of release)
- Acceleration of particle: \(10 \, \text{m/s}^2\)
### Matching with Columns:
- (A) Height of particle from ground: (p) \(40 \, \text{SI units}\)
- (B) Speed of particle: (q) \(0 \, \text{SI units}\)
- (C) Displacement of particle: (r) \(20 \, \text{SI units}\)
- (D) Acceleration of particle: (s) \(10 \, \text{SI units}\)
