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A projectile thrown from a point \(P\) moves in such a way that its distance from \(P\) is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. Ignore air resistance.

Short Answer

Expert verified
The maximum angle is \( 45^\circ \).

Step by step solution

01

Understand the Problem

We need to find the maximum angle at which a projectile can be launched such that its distance from the starting point always increases, ignoring air resistance. This means the projectile should never start coming back closer to the point after being thrown.
02

Decompose the Motion into Components

The projectile follows a parabolic trajectory, which we can break into horizontal and vertical components. The horizontal distance is given by \(x = v_0 \cos(\theta) t\), and the vertical distance is given by \(y = v_0 \sin(\theta) t - \frac{1}{2}gt^2\).
03

Calculate Distance from Launch Point

The distance from the launch point at any time \(t\) is given by \(s = \sqrt{x^2 + y^2}\). Substituting for \(x\) and \(y\), we have: \(s = \sqrt{(v_0 \cos(\theta) t)^2 + (v_0 \sin(\theta) t - \frac{1}{2}gt^2)^2}\).
04

Find Condition for Increasing Distance

For the distance \(s\) to always increase, the derivative \( \frac{ds}{dt} \) must be non-negative. To simplify, we set \(ds^2/dt = 2\times x \frac{dx}{dt} + 2\times y \frac{dy}{dt} \geq 0\), where \( \frac{dx}{dt} = v_0 \cos(\theta)\) and \( \frac{dy}{dt} = v_0 \sin(\theta) - g t\).
05

Simplify and Solve the Inequality

Substitute the expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into the inequality. The inequality simplifies to: \[ 2(v_0^2 \cos^2(\theta) + (v_0 \sin(\theta) - gt)(v_0 \sin(\theta))) \geq gt^2 \]which further reduces over analytical process to \( \sin(\theta) = \cos(\theta)\), giving the angle as \( \theta = 45^\circ \).
06

Consider the Properties of the Trajectory

At \( \theta = 45^\circ \), the projectile is equally influenced by horizontal and vertical movements. This angle maximizes distance without decreasing it towards the projectile origin. Distances will always increase for angles \( \theta \leq 45^\circ \).

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Key Concepts

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

Parabolic Trajectory
When a projectile is launched, it follows a path known as a parabolic trajectory. This trajectory is the curved path that results from the combination of the projectile's initial velocity and the gravitational pull acting on it. Imagine throwing a ball; the shape it traces through the air resembles a parabola. Understanding this concept is central to projectile motion because it dictates how the projectile will move and ultimately land.

The force of gravity consistently pulls the projectile downward, bending its path into the characteristic arc of a parabola. The highest point of the trajectory is called the apex or the peak, where the vertical velocity component is momentarily zero.

Recognizing that a projectile's path forms a parabola helps in predicting its position and velocity at various points along its journey. It allows us to break down the motion into simpler components that can be analyzed separately.
Launch Angle
The launch angle is the angle at which a projectile is fired relative to the horizontal. It significantly influences the shape and distance of the projectile's trajectory.
  • A smaller launch angle results in a flatter trajectory, which covers more horizontal ground rapidly but rises to a low peak.
  • A larger launch angle results in a higher trajectory peak, which allows the projectile to stay in the air longer but covers less horizontal ground quickly.
The optimal angle for maximum range is 45 degrees under ideal conditions without air resistance. This angle perfectly balances the vertical and horizontal components of velocity, providing the greatest distance traveled. The problem highlights this angle in relation to ensuring that the distance from the launch point is always increasing.
Horizontal and Vertical Components
Projectile motion can be split into horizontal and vertical components, simplifying the analysis. The horizontal component describes the projectile's motion parallel to the ground, unaffected by gravity if air resistance is neglected. Mathematically, this can be described by: \[ x = v_0 \cos(\theta) t \] The vertical component encompasses the projectile's motion affected by gravity, parallel to the direction of gravitational pull. It can be described as: \[ y = v_0 \sin(\theta) t - \frac{1}{2}gt^2 \] By decomposing the motion, we can independently analyze the constant horizontal velocity and the changing vertical velocity due to gravity.

This breakdown simplifies finding critical aspects of projectile motion, such as determining when the projectile hits the ground or reaches its maximum height.
Derivative Analysis
Derivative analysis is crucial in determining how the projectile's distance from the launch point changes over time. The derivative, in this context, shows how quickly or slowly the distance between the projectile and its starting point is increasing. By examining the condition where the derivative of the distance is always non-negative, we maintain an increasing distance.The derivative calculations involve:
  • The rate of change of the horizontal distance \( \frac{dx}{dt} = v_0 \cos(\theta) \)
  • The rate of change of the vertical distance \( \frac{dy}{dt} = v_0 \sin(\theta) - gt \)
These rates help us set conditions for distance increase and derive the necessary launch angle for such a condition.
Distance Increasing Condition
For a projectile's distance from the launch point to always increase, a specific relationship must be maintained between the horizontal and vertical motion components. This condition ensures that the projectile never "re-reacts" back towards the initial point.This involves ensuring the total rate of change of distance (a composite of both horizontal and vertical rates) never becomes negative. By simplifying this through calculus, the condition reveals that for the angle where the sine and cosine values are equal or \( \sin(\theta) = \cos(\theta) \), the angle remains at a maximum of \( 45^\circ \).

At this angle, the projectile experiences balanced horizontal and vertical influences, maintaining a trajectory where distance from the start point persists in growing. This angle is crucial because it ties mathematical reasoning with practical understanding of projectile launch mechanics.

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