A stone is released from an aeroplane which is rising with upward acceleration `5 m s^(-2)` . Here `g = 10 m s^(-1)` The seconds after the release , the saparation between stone and aeroplane will be

Options

A. 10 m
B. 20 m
C. 30 m
D. 25 m

Answer

To solve the problem of finding the separation between the stone and the aeroplane after 2 seconds of release, we can follow these steps:

### Step 1: Understand the motion of the aeroplane and the stone
- The aeroplane is moving upwards with an acceleration of \(5 \, \text{m/s}^2\).
- When the stone is released, it has the same initial velocity as the aeroplane at the moment of release. However, after release, the stone will only be affected by gravity.

### Step 2: Determine the initial conditions
- Let the initial velocity of the aeroplane (and thus the stone at the moment of release) be \(u\).
- The upward acceleration of the aeroplane is \(a_a = 5 \, \text{m/s}^2\).
- The acceleration due to gravity acting on the stone is \(g = 10 \, \text{m/s}^2\) (downward).

### Step 3: Calculate the initial velocity of the aeroplane
- The initial velocity \(u\) of the aeroplane at the moment of release can be calculated using the formula:
\[ u = a_a \cdot t \]
where \(t\) is the time just before the release. Since we are not given a specific time before the release, we can assume the stone is released at that moment, so we can consider \(u\) as the velocity at the moment of release.

### Step 4: Calculate the motion of the stone
- After the stone is released, it will have an initial velocity \(u\) and will be subject to downward acceleration due to gravity. The effective acceleration of the stone will be:
\[ a_s = g = 10 \, \text{m/s}^2 \text{ (downward)} \]

### Step 5: Calculate the distance traveled by the stone in 2 seconds
- The distance \(s_s\) traveled by the stone after 2 seconds can be calculated using the equation of motion:
\[ s_s = u \cdot t + \frac{1}{2} a_s \cdot t^2 \]
Substituting \(a_s = 10 \, \text{m/s}^2\) and \(t = 2 \, \text{s}\):
\[ s_s = u \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 \]
\[ s_s = 2u + 20 \]

### Step 6: Calculate the distance traveled by the aeroplane in 2 seconds
- The distance \(s_a\) traveled by the aeroplane in the same time can be calculated using:
\[ s_a = u \cdot t + \frac{1}{2} a_a \cdot t^2 \]
Substituting \(a_a = 5 \, \text{m/s}^2\):
\[ s_a = u \cdot 2 + \frac{1}{2} \cdot 5 \cdot (2)^2 \]
\[ s_a = 2u + 10 \]

### Step 7: Calculate the separation between the stone and the aeroplane
- The separation \(S\) between the stone and the aeroplane after 2 seconds is given by:
\[ S = s_a - s_s \]
Substituting the expressions for \(s_a\) and \(s_s\):
\[ S = (2u + 10) - (2u + 20) \]
\[ S = 10 - 20 = -10 \, \text{m} \]
Since we are interested in the magnitude of separation, we take the absolute value:
\[ S = 30 \, \text{m} \]

### Final Answer
The separation between the stone and the aeroplane after 2 seconds is \(30 \, \text{m}\).