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Chapter 3: Problem 55

(III) A person stands at the base of a hill that is a straight incline making an angle \(\phi\) with the horizontal (Fig. 48). For a given initial spced \(v_{0},\) at what angle \(\theta\) (to the horizontal) should objects be thrown so that the distance \(d\) they land up the hill is as large as possible?

Short Answer

Expert verified
The optimal angle is \(\theta = \frac{\pi}{4} + \frac{\phi}{2}\) for maximum distance up the hill.

Step by step solution

01

Setting up the Problem

Consider an object thrown with initial velocity \(v_0\) at an angle \(\theta\) to the horizontal. The hill makes an angle \(\phi\) with the horizontal. We aim to find the angle \(\theta\) that maximizes the distance \(d\) that the object travels up the hill.
02

Analyzing the Motion

The motion can be split into horizontal and vertical components. The initial velocity can be decomposed into horizontal \(v_{0x} = v_0 \cos \theta\) and vertical \(v_{0y} = v_0 \sin \theta\) components.
03

Equations of Motion

The equation for horizontal motion is \(x = v_{0x} t\) and for vertical motion is \(y = v_{0y} t - \frac{1}{2}gt^2\), where \(g\) is the acceleration due to gravity. At the point of landing, the object will be on the slope: \(y = x \tan \phi\).
04

Solving for Time of Flight

Substitute \(y = x \tan \phi\) in the vertical motion equation to get \(v_{0y} t - \frac{1}{2}gt^2 = v_{0x} t \tan \phi\). Solve this equation for the time of flight \(t\).
05

Expression for Distance

The distance \(d\) along the slope can be found using \(d = \sqrt{x^2 + y^2}\). By substituting \(y = x \tan \phi\), find an expression for \(d\) in terms of \(x\) and \(\phi\).
06

Optimizing the Distance

To maximize \(d\), take the derivative of the expression for \(d\) with respect to \(\theta\) and set it to zero. This involves differentiating \(d = x(1+\tan^2 \phi)^{1/2}\) with respect to \(\theta\). Solve the resulting equation for \(\theta\).
07

Simplifying the Solution

Using trigonometric identities, simplify the optimal angle \(\theta\) expression. Often, \(\theta = \frac{\pi}{4} + \frac{\phi}{2}\) is derived from these conditions for maximum range.

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Key Concepts

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

Incline Angle
A crucial component in projectile motion on an inclined plane is the incline angle, denoted by \( \phi \). This angle measures the steepness of the incline from the horizontal plain.
In our problem, the incline angle determines how curved or steep the hill is compared to a flat surface.
This impacts how far a thrown object will land on the hill. Understanding incline angles helps us see why objects on hills behave differently compared to flat ground.
  • Incline angle \( \phi \) is constant for a given problem.
  • This angle affects horizontal and vertical distances because it alters the effective gravitational pull on the projectile.
  • Knowing \( \phi \) helps us understand better angle adjustments for optimizing thrown distances.
Initial Velocity Decomposition
When studying projectile motion, the initial velocity decomposition is key. This means breaking down the starting speed \( v_0 \) into horizontal and vertical components.
This decomposition helps us understand how fast an object starts moving sideways and upwards when launched.
Each component plays a crucial role in determining how far and high the object will go.
  • Horizontal component: \( v_{0x} = v_0 \cos \theta \)
  • Vertical component: \( v_{0y} = v_0 \sin \theta \)
Understanding these components allows us to study the motion in a more straightforward way by isolating the effects of gravity and the shape of the incline.
Equations of Motion
The equations of motion describe how objects move in space and time. For projectile motion on an incline, we have different equations for horizontal and vertical directions.
These equations are essential in predicting where and when the projectile will land.
  • Horizontal Motion: \( x = v_{0x} t \)
  • Vertical Motion: \( y = v_{0y} t - \frac{1}{2}gt^2 \)
When the object hits the incline, its vertical position equals its horizontal position multiplied by \( \tan \phi \) because of the slope: \( y = x \tan \phi \).
This relationship is critical to solve for the time of flight and the exact landing spot on the hill.
Optimization of Range
Optimizing range in projectile motion means finding the best angle for the furthest distance.
For inclined surfaces, this is a bit more intricate because of the angle \( \phi \) of inclination.
The objective is to maximize distance \( d \) up the hill. To do this, we take mathematical steps:
  • Find the time of flight using motion equations.
  • Develop an expression for distance \( d \) on the incline.
  • Use calculus to find the angle \( \theta \) that gives the largest \( d \) by setting the derivative of \( d \) with respect to \( \theta \) to zero.
Upon simplifying, the optimal angle often results in \( \theta = \frac{\pi}{4} + \frac{\phi}{2} \). This angle takes into account both initial force and hill angle for maximum range.

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Most popular questions from this chapter

When Babe Ruth hit a homer over the 8.0 -m-high rightfield fence \(98 \mathrm{~m}\) from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit \(1.0 \mathrm{~m}\) above the ground and its path initially made a \(36^{\circ}\) angle with the ground.

A ball is thrown horizontally from the top of a cliff with initial speed \(v_{0}\) (at \(t=0\) ). At any moment, its direction of motion makes an angle \(\theta\) to the horizontal (Fig. \(3-47\) ). Derive a formula for \(\theta\) as a function of time, \(t,\) as the ball follows a projectile's path.

(II) A football is kicked at ground level with a spced of 18.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed \(10.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(3-52\) ). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground \((a)\) if the hot-air balloon is rising at \(5.0 \mathrm{~m} / \mathrm{s}\) relative to the ground during this throw, (b) if the hot-air balloon is descending at \(5.0 \mathrm{~m} / \mathrm{s}\) relative to the ground.

(I) A car is driven \(225 \mathrm{~km}\) west and then \(78 \mathrm{~km}\) southwest \(\left(45^{\circ}\right)\). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
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