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

Step 1: Understanding the Question:
The question provides the maximum horizontal range achievable by a projectile (bullet) and asks for its initial velocity of projection (\(u\)).
Step 2: Key Formula or Approach:
The horizontal range \(R\) of a projectile is given by \(R = \frac{u^{2} \sin(2\theta)}{g}\).
The maximum range \(R_{max}\) occurs when \(\theta = 45^{\circ}\), leading to \(\sin(2\theta) = 1\).
Thus, \(R_{max} = \frac{u^{2}}{g}\).
Step 3: Detailed Explanation:
Given: \(R_{max} = 6.4\) m.
Taking the acceleration due to gravity \(g = 10\) m/s\(^{2}\) (standard for such numerical problems unless specified otherwise).
Substitute the values into the formula:
\[ 6.4 = \frac{u^{2}}{10} \]
\[ u^{2} = 6.4 \times 10 = 64 \]
Taking the square root:
\[ u = \sqrt{64} = 8 \text{ m/s} \]
Step 4: Final Answer:
The speed of the bullet at the time of firing is \(8\) m/s. This matches option (B).