Question

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is _______ m/s. (take g = 10 m/s\(^2\))

Answer 1

Correct Answer: 8

Step 1: Understanding the Concept:
A projectile fired from the ground reaches its maximum horizontal distance (range) when the angle of projection is \( 45^\circ \). This distance is the "farthest distance" mentioned in the problem.

Step 2: Key Formula or Approach:
Maximum Horizontal Range: \[ R_{\text{max}} = \frac{u^2}{g} \]

Step 3: Detailed Explanation:
Given: Maximum distance \( R_{\text{max}} = 6.4 \text{ m} \) and \( g = 10 \text{ m/s}^2 \).
We know that for \( \theta = 45^\circ \), \( R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} \).
Substituting the values:
\[ 6.4 = \frac{u^2}{10} \]
\[ u^2 = 64 \]
\[ u = \sqrt{64} = 8 \text{ m/s} \]

Step 4: Final Answer:
The speed of the bullets is 8 m/s.