The maximum range of a gun on horizontal terrain is 25 km. If g= 10 `ms^(-2)`, the muzzle velocity of the shell

Options

A. `250 ms^(-1)`
B. `2500 ms^(-1)`
C. `25 ms^(-1)`
D. `500 ms^(-1)`

Explanation

To find the muzzle velocity of the shell given the maximum range of a gun on horizontal terrain, we can use the formula for the range of a projectile. The maximum range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( u \) is given by:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

Where:
- \( R \) is the range,
- \( u \) is the initial velocity (muzzle velocity),
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of projection.

### Step 1: Identify the maximum range condition
The maximum range occurs when \( \sin(2\theta) \) is maximized. The maximum value of \( \sin(2\theta) \) is 1, which occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).

### Step 2: Substitute the values into the range formula
Given:
- Maximum range \( R = 25 \) km = \( 25 \times 1000 \) m = \( 25000 \) m,
- \( g = 10 \, \text{m/s}^2 \).

Using the maximum range formula:

\[ R = \frac{u^2}{g} \]

Substituting the known values:

\[ 25000 = \frac{u^2}{10} \]

### Step 3: Solve for \( u^2 \)
To find \( u^2 \), we can rearrange the equation:

\[ u^2 = 25000 \times 10 \]

\[ u^2 = 250000 \]

### Step 4: Calculate \( u \)
Now, take the square root to find \( u \):

\[ u = \sqrt{250000} \]

\[ u = 500 \, \text{m/s} \]

### Final Answer
The muzzle velocity of the shell is \( 500 \, \text{m/s} \).

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