Problem 40 - Master Resource Book in JEE Main Physics

Question

A large number of bullets are fired in the all directions with same speed \(v\). What is the maximum area on the ground on which these bullets will spread? (a) \(\pi \frac{v^{2}}{g}\) (b) \(\pi \frac{v^{4}}{g^{2}}\) (c) \(n^{2} \frac{v^{4}}{g^{2}}\) (d) \(\pi^{2} \frac{v^{2}}{g^{2}}\)

Step-by-Step Solution

Answer

The maximum area is (b) \(\pi \frac{v^{4}}{g^{2}}\).

Step 1: Understand the Problem

We are given that bullets are fired in all directions with the same speed, and we need to determine the maximum area on the ground that they will cover. The bullets behave like projectiles, so we must use projectile motion concepts.

Step 2: Recall Projectile Motion Range Formula

For a projectile, the horizontal range is given by the formula \(R = \frac{v^2 \sin(2\theta)}{g}\), where \(\theta\) is the launch angle and \(g\) is the acceleration due to gravity.

Step 3: Determine Maximum Range

The range is maximized when \(\sin(2\theta) = 1\), which occurs when \(2\theta = 90^\circ\) or \(\theta = 45^\circ\). Thus, the maximum horizontal range \(R_{max} = \frac{v^2}{g}\).

Step 4: Geometry of Maximum Area Coverage

Since bullets are fired in all directions covering a circular region, the area covered is a circle with radius equal to the maximum range \(R_{max} = \frac{v^2}{g}\). The area \(A\) of a circle is \(\pi r^2\).

Step 5: Calculate Maximum Area

Substitute \(r = R_{max}\) into the circle area formula: \[A = \pi \left(\frac{v^2}{g}\right)^2\]. Simplifying this gives: \[A = \pi \frac{v^4}{g^2}\].

Key Concepts

Horizontal Range

The concept of horizontal range is fundamental in understanding projectile motion. It refers to the total horizontal distance covered by a projectile. In projectile motion, an object is launched into the air and its horizontal and vertical motions occur simultaneously but independently. The horizontal range is calculated using the formula:

• \( R = \frac{v^2 \sin(2\theta)}{g} \)

Here, \(v\) is the initial velocity, \(\theta\) is the angle of launch, and \(g\) represents the acceleration due to gravity, roughly equal to 9.81 m/s² on Earth's surface.
The range of the projectile depends significantly on the angle of launch and the velocity. For instance, if launched at 45 degrees, it reaches its maximum range. This occurs because \(\sin(90^\circ) = 1\), optimizing the range formula. Thus, understanding this formula helps predict how far a projectile will land from its starting point.

Maximum Range

The maximum range of a projectile is the furthest extent it can travel horizontally. To achieve this, the launch angle \(\theta\) must be precisely 45 degrees. In this scenario, the projectile is neither too steep nor too flat, which optimizes the range due to an optimal balance between vertical and horizontal components.
When launched at this angle, the sine of double the launch angle is \( \sin(90^\circ) \), which is 1. Thus, the maximum horizontal range can be expressed by the simplified formula:

• \( R_{\text{max}} = \frac{v^2}{g} \)

Understanding the relationship between the launch angle and range can be intuitive. At 45 degrees, a projectile efficiently converts its initial velocity into equal horizontal and vertical motions, maximizing its travel distance. This knowledge is particularly useful in various real-world applications, such as sports and engineering, where precise range estimations are vital.

Projectile Motion Formula

Projectile motion encompasses both the horizontal and vertical movements of an object. It's essential to understand the formulas involved to fully predict the trajectory of a projectile. The key equations describe:

• Horizontal motion: \( x = vt \cos(\theta) \)
• Vertical motion: \( y = vt \sin(\theta) - \frac{1}{2}gt^2 \)

Hovering over all trajectories, the projectile follows a parabolic path, dictated by these equations. For any initial speed and launch angle, these formulas help calculate the range, maximum height, and time of flight.
In the case of maximum area coverage (as in the exercise), the projectile trajectory forms a circle when fired in all directions, with the radius given by the maximum range \( R_{\text{max}} \). Using the range and maximum area formula, the area becomes \( A = \pi \left(\frac{v^2}{g}\right)^2 \), which signifies that calculating projectile motion is integral to determining the spread of range on the ground, expanding practical applications from physics to diverse fields.