Consider a shell that has a muzzle velocity of `45ms^(-1)` fired from the tail gun of an airplane moving horizontally with a velocity of `215 ms^(-1)` . The tail gun can be directed at any angle with the vertical in the plane of motion of the airplane. The shell is fired when the plane is above point A on ground, and the plane is above point B on ground when the shell hits the ground. (Assume for simplicity that the Earth is flat)

Options

A. Shell may hit the ground at point A.
B. Shell may hit the ground at point B.
C. Shell may hit a point on earth which is behind point A.
D. Shell may hit a point on earth which is behind point B.

Explanation

To solve the problem, we need to analyze the motion of the shell fired from the airplane. The shell has a muzzle velocity of \(45 \, \text{m/s}\) and the airplane is moving horizontally at \(215 \, \text{m/s}\). We will consider the horizontal and vertical motions separately.

### Step 1: Determine the time of flight of the shell
The vertical motion of the shell is influenced by gravity. The time of flight \(t\) can be calculated using the formula for the vertical motion under gravity:

\[ h = \frac{1}{2} g t^2 \]

where \(g \approx 9.81 \, \text{m/s}^2\) is the acceleration due to gravity. Rearranging this gives:

\[ t = \sqrt{\frac{2h}{g}} \]

### Step 2: Calculate the horizontal distance traveled by the shell
The horizontal distance \(d\) traveled by the shell while it is in the air can be calculated using the horizontal velocity of the shell. The horizontal component of the shell's velocity is the sum of the airplane's velocity and the horizontal component of the shell's muzzle velocity.

If the shell is fired at an angle \(\theta\) with respect to the horizontal, the horizontal component of the shell's velocity \(V_{x}\) is:

\[ V_{x} = V_{muzzle} \cdot \cos(\theta) + V_{airplane} \]

The distance traveled horizontally is given by:

\[ d = V_{x} \cdot t \]

### Step 3: Relate the distances to points A and B
Let’s denote the distance from point A to point B as \(D\). The airplane travels this distance while the shell is in the air:

\[ D = V_{airplane} \cdot t \]

### Step 4: Set up the equations
From the above steps, we have two equations:

1. \(t = \sqrt{\frac{2h}{g}}\)
2. \(D = V_{airplane} \cdot t\)

Substituting the expression for \(t\) from the first equation into the second gives:

\[ D = V_{airplane} \cdot \sqrt{\frac{2h}{g}} \]

### Step 5: Solve for the height \(h\)
To find the height \(h\), we can rearrange the equation:

\[ h = \frac{gD^2}{2V_{airplane}^2} \]

### Conclusion
The shell will hit the ground at a point that is horizontally displaced from point A, and depending on the angle of firing, it may land before or after point B.