Chapter 3: Problem 85
\(\Lambda\) 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) \(\pi\left(\frac{u^{2}}{g}\right)^{2}\) (b) \(\pi\left(\frac{u^{2}}{2 g}\right)^{2}\) (c) \(\pi\left(\frac{u}{g}\right)^{2}\) (d) \(\pi\left(\frac{u}{2 g}\right)^{2}\)
Short Answer
Step by step solution
Understand the Problem
Identify Key Parameters
Determine Maximum Range
Calculate Maximum Area Covered
Identify the Correct Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Range of Projectile
- The initial speed or velocity at which the object is launched
- The angle of launch relative to the ground
- Acceleration due to gravity (usually constant at approximately 9.8 m/s² on Earth)
Maximum Range
Projectile Angle
- A 45-degree angle grants the maximum range for a given speed and no air resistance.
- Angles steeper than 45 degrees will result in a higher but shorter path.
- Shallower angles will cover more horizontal distance quickly but won't travel as far overall.
Horizontal Range Formula
- \( R \) is the horizontal range.
- \( u \) is the initial velocity.
- \( \theta \) is the launch angle from the horizontal.
- \( g \) is the acceleration due to gravity (9.8 m/s² on Earth).