Question

A number of bullets are fired in all possible directions with the same initial velocity u. The maximum area of ground covered by bullets is

\[A.\,\pi {{\left( \dfrac{{{u}^{2}}}{g} \right)}^{2}}\]
\[B.\,\pi {{\left( \dfrac{{{u}^{2}}}{2g} \right)}^{2}}\]
\[C.\,\pi {{\left( \dfrac{u}{g} \right)}^{2}}\]

Answer 1

Hint: The formula for calculating the range of a projectile motion should be used to solve this problem, as the type of motion a fired bullet follows will be a projectile motion. As, the bullets are fired in all the directions, so, they form a circle with the radius being equal to the range of the projectile. Thus, the maximum area covered by the fired bullets is calculated as the area of the circle with the radius being the maximum range.

Formula used:
\[R=\dfrac{{{u}^{2}}\sin (2\theta )}{g}\]

Complete step-by-step answer:
The formula for calculating the range of the projectile motion is given as follows.
\[R=\dfrac{{{u}^{2}}\sin (2\theta )}{g}\]
Where u is the initial speed of the object, g is the gravitational constant and \[\theta \]is the angle made by the projection of the object with the horizontal.
In order to cover the maximum area, the range of the projection should also be maximum. Thus, we need to find the value of the maximum range. And, the range will be maximum when the angle will be equal to \[45{}^\circ \], because, \[\sin (2\times 45{}^\circ )=1\], the maximum value of the sin function.
Therefore, the maximum range is,
\[\begin{align}
  & R=\dfrac{{{u}^{2}}\sin (2\times 45{}^\circ )}{g} \\
 & \Rightarrow R=\dfrac{{{u}^{2}}\sin (90{}^\circ )}{g} \\
 & \Rightarrow R=\dfrac{{{u}^{2}}}{g} \\
\end{align}\]
Now let us compute the area of the circle, as the bullets are fired in all the directions, so, they form a circle with radius being equal to the range of the projectile.
The area covered is given as follows.
\[A=\pi {{R}^{2}}\]
Substitute the expression of the range in the above equation, as the radius of the circle equals the range of the projectile.
\[A=\pi {{\left( \dfrac{{{u}^{2}}}{g} \right)}^{2}}\]
As, the maximum area of the ground covered by the fired bullets equals \[\pi {{\left( \dfrac{{{u}^{2}}}{g} \right)}^{2}}\]

So, the correct answer is “Option A”.

Note: The main points to remember while solving such problems are, the area covered in all the directions will be the area of the circle. the range of the projectile will be maximum, only when the angle made by the projected object with the horizontal equals \[45{}^\circ \], because, \[\sin (2\times 45{}^\circ )=1\], the maximum value of the sin function.