Speed of Bullets Fired from Ground with Maximum Range

To find the speed of the bullets fired from the gun, we consider the equation for the horizontal range of a projectile fired from ground level.

The horizontal range $R$ of a projectile fired with an initial speed $u$ at an angle of projection $\theta$ with the horizontal is given by: $R = \frac{u^2 \sin(2\theta)}{g}$

The farthest distance (maximum horizontal range $R_{\text{max}}$) is achieved when the angle of projection is $\theta = 45^\circ$, which gives $\sin(2\theta) = \sin(90^\circ) = 1$.

Therefore, the maximum range is: $R_{\text{max}} = \frac{u^2}{g}$

Given values:

• Maximum range, $R_{\text{max}} = 6.4\text{ m}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

Substituting the given values into the formula: $6.4 = \frac{u^2}{10}$

Solving for $u^2$: $u^2 = 6.4 \times 10 = 64$

Taking the square root on both sides: $u = \sqrt{64} = 8\text{ m/s}$

Thus, the speed of the bullets from the gun is $8\text{ m/s}$.