A large number of bullets are fired in all directions with same speed u. What is the maximum area on the ground on which these bullets will spread

The correct option is: πu⁴/g².

To find the maximum area on the ground where the bullets can spread, we need to determine the maximum horizontal distance (or maximum range) that any single bullet can travel when fired with an initial speed $u$.

The horizontal range $R$ of a projectile launched at an angle $\theta$ to the horizontal with an initial speed $u$ is given by the formula:

$R=\frac{u^{2}sin\left(2\theta \right)}{g}$

where $g$ is the acceleration due to gravity.

The horizontal range is maximized when the term $sin\left(2\theta \right)$ reaches its maximum value of 1, which occurs at a launch angle of $\theta ={45}^{\circ }$. Thus, the maximum range $R_{max}$ is:

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

Since the bullets are fired in all horizontal directions (360 degrees around the launch point), the area over which they spread forms a circle on the ground. The boundary of this circle is determined by the maximum horizontal distance a bullet can travel. Therefore, the radius of this circular area is equal to the maximum range $R_{max}$.

The maximum area $A_{max}$ of this circular region is calculated using the formula for the area of a circle:

$A_{max}=\pi R_{max}^2$

Substituting the value of $R_{max}$ into the area equation:

$A_{max}=\pi {\left(\frac{u^2}{g}\right)}^2$

Simplifying the expression, we get:

$A_{max}=\frac{\pi u^4}{g^2}$

Thus, the maximum area on the ground on which the bullets will spread is $\frac{\pi u^4}{g^2}$, which corresponds to the option πu⁴/g².