Question

The maximum range of a gun on horizontal plane is 16 km. If g = 10ms^-2 , then muzzle velocity of a shell is

• A. 160 ms^-1
• B. 200\(\sqrt2\) ms^-1
• C. 400 ms^-1
• D. 800 ms^-1

Answer 1

The Correct Option is C

1. Understand the Given Data:

We are given:

• The maximum range of the projectile, $ R = 16 \, \text{km} = 16000 \, \text{m} $,
• The acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $.

The goal is to find the muzzle velocity $ v_0 $ of the shell.

2. Use the Formula for Maximum Range of a Projectile:

The maximum range of a projectile is given by the formula:

$$ R = \frac{v_0^2}{g} $$

Where:

• $ R $: Maximum range,
• $ v_0 $: Muzzle velocity,
• $ g $: Acceleration due to gravity.

3. Rearrange the Formula to Solve for $ v_0 $:

We can solve for $ v_0 $ as follows:

$$ v_0^2 = R \times g $$

Taking the square root of both sides:

$$ v_0 = \sqrt{R \times g} $$ 

4. Substitute the Given Values:

Substitute $ R = 16000 \, \text{m} $ and $ g = 10 \, \text{m/s}^2 $ into the formula:

$$ v_0 = \sqrt{16000 \, \text{m} \times 10 \, \text{m/s}^2} $$ $$ v_0 = \sqrt{160000} $$ 

5. Calculate the Value of $ v_0 $:

Taking the square root of $ 160000 $:

$$ v_0 = 400 \, \text{m/s} $$ 

6. Final Answer:

The muzzle velocity of the shell is $ 400 \, \text{m/s} $.

Answer 2

The maximum range of a projectile on a horizontal plane is given by the formula:

$R_\text{max} = \frac{u^2}{g}$

where:

• $R_\text{max}$ is the maximum range,
• $u$ is the initial velocity (muzzle velocity in this case), and
• $g$ is the acceleration due to gravity.

We are given $R_\text{max} = 16\,\text{km} = 16000\,\text{m}$ and $g = 10\,\text{m/s}^2$. We need to find $u$.

Rearranging the formula to solve for $u$:

$u = \sqrt{R_\text{max} \cdot g}$

Substituting the given values:

$u = \sqrt{16000\,\text{m} \times 10\,\text{m/s}^2} = \sqrt{160000\,\text{m}^2/\text{s}^2} = 400\,\text{m/s}$

The correct answer is (C) 400 ms^-1.