The maximum range of a gun on horizontal terrain is 1km. If `g = 10 ms^(-2)`, what must be the muzzle velocity of the shell ?

Options

A. `400 ms^(-1)`
B. `200 ms^(-1)`
C. `100 ms^(-1)`
D. `50 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 \( R \) is given by the equation:

\[ R = \frac{u_0^2 \sin(2\theta)}{g} \]

Where:
- \( R \) is the range,
- \( u_0 \) is the muzzle velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Identify the maximum range and gravitational acceleration
From the problem, we know:
- The maximum range \( R = 1 \text{ km} = 1000 \text{ m} \)
- The gravitational acceleration \( g = 10 \text{ m/s}^2 \)

### Step 2: Determine the angle for maximum range
The maximum range occurs when the angle of projection \( \theta \) is \( 45^\circ \). At this angle, \( \sin(2\theta) = \sin(90^\circ) = 1 \).

### Step 3: Substitute values into the range formula
Using the maximum range formula, we can rewrite it as:

\[ R = \frac{u_0^2}{g} \]

Substituting the known values:

\[ 1000 = \frac{u_0^2}{10} \]

### Step 4: Solve for \( u_0^2 \)
To isolate \( u_0^2 \), multiply both sides by \( g \):

\[ u_0^2 = 1000 \times 10 \]
\[ u_0^2 = 10000 \]

### Step 5: Calculate \( u_0 \)
Now, take the square root of both sides to find \( u_0 \):

\[ u_0 = \sqrt{10000} \]
\[ u_0 = 100 \text{ m/s} \]

### Conclusion
The muzzle velocity of the shell must be \( 100 \text{ m/s} \).

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