An aeroplane is flying vertically upwards. When it is at a height of `1000 m` above the ground and moving at a speed of `367 m//s`., a shot is fired at it with a speed of 567 `ms^(-1)` from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit?

Options

A. `gt5ms^(-2)`
B. `gt10 ms^(-2)`
C. `lt10 ms^(-2)`
D. Not possible

Explanation

To solve the problem, we need to determine the acceleration of the airplane (denoted as \( a_P \)) required to escape being hit by the shot fired from below. Let's break down the solution step by step.

### Step 1: Understand the scenario
The airplane is flying vertically upwards at a height of 1000 m with an initial speed of \( v_P = 367 \, \text{m/s} \). A shot is fired from the ground with a speed of \( v_S = 567 \, \text{m/s} \). We need to find the required acceleration of the airplane so that it can escape being hit by the shot.

### Step 2: Define the relative motion
The shot is fired directly upwards. To determine the conditions under which the airplane escapes being hit, we will analyze the relative motion between the airplane and the shot.

### Step 3: Establish the equations of motion
1. The distance between the airplane and the point of firing is \( S = 1000 \, \text{m} \).
2. The initial velocity of the shot relative to the airplane is given by:
\[ u_{relative} = v_S - v_P = 567 \, \text{m/s} - 367 \, \text{m/s} = 200 \, \text{m/s} \]
3. The acceleration of the shot is \( -g \) (downward), where \( g \approx 9.81 \, \text{m/s}^2 \).

### Step 4: Apply the equation of motion
Using the equation of motion for relative velocity:
\[ v_{relative}^2 = u_{relative}^2 + 2 \cdot a_{relative} \cdot S \]
We want the final relative velocity \( v_{relative} \) to be zero when the shot reaches the height of the airplane, thus:
\[ 0 = (200)^2 + 2 \cdot (a_P - g) \cdot 1000 \]
This simplifies to:
\[ 0 = 40000 + 2000(a_P - g) \]

### Step 5: Solve for \( a_P \)
Rearranging the equation gives:
\[ 2000(a_P - g) = -40000 \]
\[ a_P - g = -20 \]
\[ a_P = g - 20 \]

### Step 6: Substitute the value of \( g \)
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ a_P = 9.81 - 20 = -10.19 \, \text{m/s}^2 \]
Since we need the airplane to have a positive acceleration to escape, we need:
\[ a_P > 10 \, \text{m/s}^2 \]

### Conclusion
The required acceleration of the airplane to escape being hit is:
\[ a_P > 10 \, \text{m/s}^2 \]