The maximum range of a gun from horizontal terrain is `16 km`. If `g = 10 m//s^(2)` what must be the muzzle velocity of the shell?

Options

A. `160ms^(-1)`
B. `200sqrt(2)ms^(-1)`
C. `400ms^(-1)`
D. `800ms^(-1)`

Answer

To find the muzzle velocity of the shell given the maximum range, 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 the equation:

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

For maximum range, the angle \( \theta \) should be \( 45^\circ \). At this angle, \( \sin 2\theta = \sin 90^\circ = 1 \). Therefore, the equation simplifies to:

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

### Step 1: Rearranging the formula
We need to rearrange the formula to solve for \( u \):

\[ u^2 = R \cdot g \]

### Step 2: Substituting the values
We know:
- The maximum range \( R = 16 \, \text{km} = 16000 \, \text{m} \) (since 1 km = 1000 m)
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Now, substituting these values into the equation:

\[ u^2 = 16000 \, \text{m} \cdot 10 \, \text{m/s}^2 \]

### Step 3: Calculating \( u^2 \)
Calculating the right side:

\[ u^2 = 160000 \, \text{m}^2/\text{s}^2 \]

### Step 4: Taking the square root
Now, we find \( u \) by taking the square root of both sides:

\[ u = \sqrt{160000} \, \text{m/s} \]

Calculating the square root:

\[ u = 400 \, \text{m/s} \]

### Conclusion
Thus, the muzzle velocity of the shell must be \( 400 \, \text{m/s} \).

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