Question

A large number of bullets are fired in all directions with the same speed $u$. What is the maximum area on the ground on which these bullets will spread?
A.)$\pi\;\dfrac{u^2}{g}$
B.)$\pi\;\dfrac{u^4}{g^2}$
C.)$\pi^2\;\dfrac{u^4}{g^2}$
D.)$\pi^2\;\dfrac{u^2}{g^2}$

Answer 1

Try and interpret the problem as a case of projectile motion. The projectiles together form a circular distribution on the ground. The radius of this circle will be the maximum range of the projectile, i.e., the launch angle of the projectile $\theta = 45^{\circ}$.

Formula used:
Area of circle = $\pi \; r^2$, where r is the radius of the circle.
Range of a projectile $R = \dfrac{u^2\;sin\;2\theta}{g}$ where u is the launch velocity, $\theta$ is the launch angle, and g is the acceleration due to gravity.

Complete answer:
Let us deconstruct the question to understand it in terms that are familiar to us.
The question basically deals with a case of projectile motion. Projectile motion is a form of motion where an object moves in a parabolic path. It occurs when there is one force that launches the object after which the only influencing force is gravity. The path that the object follows under this influence of gravity is called projectile trajectory. In most cases for our convenience we neglect air resistance.
A bullet fired from a gun is an example of projectile motion.
The range of a projectile can be understood as the horizontal distance between the launch point and the point where the projectile hits the ground.

Now, let us try and understand what the question entails.
We have a gun that fires bullets in all directions. This means that the distribution of all bullets landing on the ground would be in the form of a circle, since they’re all fired with the same launch velocity $u$, and the radius of this circle would be equal to the range of the projectile.

Therefore, the area covered by the bullets on the ground can be given as $A = \pi R^2$
The range of a projectile is given as $R = \dfrac{u^2\;sin\;2\theta}{g}$.
Now, the question asks for us to find the “maximum” area that can be obtained from the projectile distribution. Therefore, we need the maximum range that the projectiles can achieve.
From the equation for range $R$ we see that the maximum value of R that we can get is

$R_{max} = \dfrac{u^2}{g}$ where the value of $sin\;2\theta = 1 \Rightarrow \theta = 45^{\circ}$
Substituting $R_{max}$ in area, we get:
$A_{max} = \pi \times R^{2}_{max} \Rightarrow \pi \;\dfrac{u^4}{g^2}$

So, the correct answer is “Option B”.

Note:
Here, we have assumed that the bullets are launched from the flat ground. However, if you should encounter any projectile that is launched from a certain height $h$, then the range of the projectile takes the form:
$R = \dfrac{u^2}{2g} \left(1 +\sqrt{ 1 + \dfrac{2gh}{u^2\;sin\theta}} \right)\;sin\;2\theta$