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Show that a projectile achieves its maximum range when it is fired at \(45^{\circ}\) above the horizontal if \(y=y_{0}\).

Short Answer

Expert verified
Maximum range occurs at a launch angle of 45° if initial and final heights are the same.

Step by step solution

01

Understand the Problem

In projectile motion, we need to find the angle at which the projectile achieves maximum horizontal range. Given that the initial and final vertical positions are the same \(y_0\), the projectile lands at the same horizontal level from which it was launched.
02

Write the Range Formula

The range of a projectile is given by the formula: \[ R = \frac{v^2_0 \sin(2\theta)}{g} \]Here, \(v_0\) is the initial velocity, \(\theta\) is the angle of launch, and \(g\) is the acceleration due to gravity.
03

Analyze the Sin Function

The \sin(2\theta)\ function reaches its maximum value of 1 when \(2\theta = 90^{\circ}\), hence \(\theta = 45^{\circ}\). This is key for maximizing the product in the range equation.
04

Conclusion

Substituting \(\theta = 45^{\circ}\) into the range formula, we maximize the range \(R\). Therefore, a projectile achieves its maximum range when fired at \(45^{\circ}\) when the launch and landing heights are equal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum Range
In the realm of projectile motion, achieving the maximum range is often a critical goal. The range of a projectile, which refers to the horizontal distance it covers from launch to landing, depends on several factors
  • Initial speed
  • Launch angle
  • Effect of gravity
Among these, the launch angle plays a pivotal role in maximizing the range.
When the initial and final vertical positions of a projectile are the same, it means the projectile lands at the same height it was fired from. Under these circumstances, the maximum range is achieved when the launch angle is
exactly at 45 degrees. This angle provides the perfect balance between horizontal velocity and vertical velocity, ensuring that the projectile covers the greatest possible horizontal distance. Understanding and applying this principle is essential in fields ranging from military science to sports dynamics. Achieving the maximum range widens the application efficiency remarkably across various practical scenarios.
Launch Angle
The launch angle of a projectile substantially impacts its trajectory and range. Picture a projectile being launched at various angles:
  • A steep angle gives a higher flight path with less horizontal reach
  • A shallow angle yields a longer flight path with lower maximum height
The magical angle of 45 degrees serves a special role in projectile motion.
At this angle, the projectile experiences an ideal balance of vertical and horizontal velocity components. This creates a perfect trajectory where neither height nor distance is compromised, thus optimizing the range. The balance of sine and cosine components at 45 degrees ensures that both height and horizontal motion contribute equally to the overall motion of the projectile.
Moreover, for any given initial speed, this angle reduces air resistance and gravitational effects, making the projectile's path as efficient as possible in ideal, frictionless conditions.
Range Formula
To understand and calculate projectile range, the range formula is indispensable. It is given by:\[R = \frac{v^2_0 \sin(2\theta)}{g}\]

Components of the Range Formula

This formula integrates several critical elements that determine the range:
  • \(v_0\): The initial velocity of the projectile. Higher initial speeds contribute to longer ranges.
  • \(\theta\): The launch angle. As established, a 45-degree angle maximizes range.
  • \(g\): Acceleration due to gravity. Typically, this is around 9.81 m/s² on Earth.
The function \(\sin(2\theta)\) plays a key role here. This trigonometric component varies with the launch angle, reaching its peak at a value of 1 when \(2\theta = 90^\circ\), corresponding to a launch angle \(\theta = 45^\circ\).
This elegant balance results in the maximum possible multiplication factor for the formula, allowing the projectile to achieve a significant horizontal distance.
Horizontal Range
The horizontal range of a projectile is the total horizontal distance it travels while in motion. This concept is crucial when discussing projectile motion because it sums up the effective reach of the projectile.
Horizontal range is directly tied to:
  • Initial speed \(v_0\)
  • Launch angle \(\theta\)
  • Gravity's pull \(g\)
In frictionless environments, where air resistance is negligible, achieving maximum horizontal range involves optimizing these factors.
At a launch angle of 45 degrees, which evenly distributes the velocity between upward and forward momentum, the horizontal range is maximized. This reveals a practical insight into designing experiments, sports moves, and even planning safety measures in real-life applications that require projectile motion considerations. By ensuring that these factors are fine-tuned, you can achieve an optimal horizontal range in your project.

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Most popular questions from this chapter

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