Question

Bullets are fired from a gun in all the direction at all angles. Maximum range on horizontal ground is \(6.4\) m. Find speed of bullet at the time of firing.

• A. \(4\sqrt{2}\) m/s
• B. \(8\) m/s
• C. \(16\) m/s
• D. \(8\sqrt{2}\) m/s

Hint:

Remember that maximum range is achieved at an angle of \(45^{\circ}\) and is exactly four times the maximum height achieved at that same angle (\(R_{max} = 4 H_{max}\) at \(45^{\circ}\)). Always check if using \(g = 9.8\) or \(g = 10\) gives a perfect square; for \(6.4\), \(g=10\) is clearly intended.

Answer 1

The Correct Option is B

To determine the speed of the bullet at the time of firing, we can use the formula for the maximum range of a projectile. The maximum range \( R_{\text{max}} \) is given by:

\(R_{\text{max}} = \frac{v^2}{g}\)

where:

• \(R_{\text{max}}\) is the maximum range,
• \(v\) is the initial velocity (or speed) of the projectile,
• \(g\) is the acceleration due to gravity, approximately \(10 \, \text{m/s}^2\).

Given that the maximum range \(R_{\text{max}}\) is \(6.4\) m, we can substitute the values into the formula and solve for \(v\):

\(6.4 = \frac{v^2}{10}\)

Multiply both sides by \(9.8\) to isolate \(v^2\):

\(v^2 = 6.4 \times 10\)

Calculate the right-hand side:

\(v^2 = 64\)

Taking the square root of both sides gives:

\(v = \sqrt{64}\)

Calculating this, we find:

\(v = 8 \, \text{m/s}\)

Therefore, the correct answer is 8 m/s.