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Chapter 4: Problem 47

(a) A ball is thrown straight up with velocity \(v\). Find an equation for the height to which it rises.(answer check available at lightandmatter.com) (b) Generalize your equation for a ball thrown at an angle \(\theta\) above horizontal, in which case its initial velocity components are \(v_{x}=v \cos \theta\) and \(v_{y}=v \sin \theta\).(answer check available at lightandmatter.com)

Short Answer

Expert verified
(a) \( h = \frac{v^2}{2g} \); (b) \( h = \frac{v^2 \sin^2 \theta}{2g} \).

Step by step solution

01

Understanding the Problem

We are given a ball thrown straight up with an initial velocity \( v \) and need to find the maximum height it reaches. For part (b), generalize this for a ball thrown at an angle \( \theta \).
02

Apply the Vertical Motion Formula for Part (a)

When a ball is thrown straight up, its motion can be described by the equation for maximum height: \( h = \frac{v^2}{2g} \), where \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏). This comes from setting the final velocity at maximum height to 0 and solving the kinematic equation \( v_f^2 = v_i^2 - 2gh \).
03

Convert the Velocity to Vertical Component for Part (b)

When thrown at an angle \( \theta \), the vertical component of velocity is \( v_y = v \sin \theta \). This component will influence the maximum height reached.
04

Calculate the Maximum Height for Part (b)

Substitute the vertical velocity component into the height formula: \( h = \frac{(v \sin \theta)^2}{2g} \). Simplifying gives \( h = \frac{v^2 \sin^2 \theta}{2g} \). This equation represents the maximum height for a ball thrown at angle \( \theta \).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It involves parameters such as position, velocity, and acceleration. In projectile motion, which is a type of kinematics, an object moves along a curved path under the influence of gravity alone, after an initial force has been applied.

This motion can be analyzed independently in the horizontal and vertical directions. Both directions are components of motion that work together, but they do not affect each other. This makes the analysis simpler and allows us to use basic equations of motion to solve complex problems, like a ball being thrown at an angle.
Vertical Motion
Vertical motion is the part of kinematics that deals with motion along the vertical axis. For an object thrown upwards, it moves against gravity, slowing down until it reaches its highest point where the velocity becomes zero.

In the case of a projectile thrown vertically, the initial velocity is positive, and as the object rises, gravity, which acts downwards, reduces its speed until it stops momentarily at the maximum height. After reaching this point, the object will start to descend, accelerating back towards the Earth due to gravitational pull.
Maximum Height
Maximum height is an important concept in projectile motion and vertical motion. It is the highest point an object will reach when thrown upwards. At this peak, the vertical component of the velocity is zero.

We use the formula to determine the maximum height: \[h = \frac{v^2}{2g}\]where \(h\) is the height, \(v\) is the initial velocity, and \(g\) is the acceleration due to gravity (approximately 9.8 m/s虏). The formula tells us how the maximum height is related to both the initial velocity and the gravitational force. For an object thrown at an angle, the vertical component of the velocity is used in the formula to calculate how high the object will go.
Initial Velocity Components
The initial velocity of a projectile is often divided into two components: horizontal and vertical. These components are essential to understanding projectile motion.

When a projectile is launched at an angle \(\theta\) from the horizontal, the initial velocity \(v\) has:
  • A horizontal component: \(v_x = v \cos \theta\)
  • A vertical component: \(v_y = v \sin \theta\)

These components allow us to analyze the projectile's motion separately in the horizontal and vertical directions to predict its trajectory accurately.
Acceleration due to Gravity
Acceleration due to gravity is a crucial factor in projectile motion. It is the constant acceleration that acts on any object moving under the influence of gravity alone. For our purposes, it's approximately \(9.8 \text{ m/s}^2\) directed towards the Earth.

This acceleration affects the vertical motion of the projectile. While the horizontal component of motion stays consistent in the absence of air resistance, the vertical motion is constantly affected by gravity, causing deceleration as the object rises and acceleration as it falls back down.

Recognizing how gravity affects different components of motion is key to solving projectile motion problems effectively.

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Most popular questions from this chapter

If you walk \(35 \mathrm{~km}\) at an angle \(25^{\circ}\) counterclockwise from east, and then \(22 \mathrm{~km}\) at \(230^{\circ}\) counterclockwise from east, find the distance and direction from your starting point to your destination. (answer check available at lightandmatter.com)

Two wheels of radius \(r\) rotate in the same vertical plane with angular velocities \(+\Omega\) and \(-\Omega\) about axes that are parallel and at the same height. The wheels touch one another at a point on their circumferences, so that their rotations mesh like gears in a gear train. A board is laid on top of the wheels, so that two friction forces act upon it, one from each wheel. Characterize the three qualitatively different types of motion that the board can exhibit, depending on the initial conditions.

In a well known stunt from circuses and carnivals, a motorcyclist rides around inside a big bowl, gradually speeding up and rising higher. Eventually the cyclist can get up to where the walls of the bowl are vertical. Let's estimate the conditions under which a running human could do the same thing. (a) If the runner can run at speed \(v\), and her shoes have a coefficient of static friction \(\mu_{s}\), what is the maximum radius of the circle?(answer check available at lightandmatter.com) (b) Show that the units of your answer make sense. (c) Check that its dependence on the variables makes sense. (d) Evaluate your result numerically for \(v=10 \mathrm{~m} / \mathrm{s}\) (the speed of an olympic sprinter) and \(\mu_{s}=5\). (This is roughly the highest coefficient of static friction ever achieved for surfaces that are not sticky. The surface has an array of microscopic fibers like a hair brush, and is inspired by the hairs on the feet of a gecko. These assumptions are not necessarily realistic, since the person would have to run at an angle, which would be physically awkward.)(answer check available at lightandmatter.com)

Derive a formula expressing the kinetic energy of an object in terms of its momentum and mass.(answer check available at lightandmatter.com)

A mass \(m\) moving at velocity \(v\) collides with a stationary target having the same mass \(m\). Find the maximum amount of energy that can be released as heat and sound. (answer check available at lightandmatter.com)
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