/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 \- Wes and Lindsay stand on the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\- Wes and Lindsay stand on the roof of a building. Wes leans over the edge and drops an apple. Lindsay waits \(1.25 \mathrm{~s}\) after Wes releases his fruit and throws an orange straight down at \(28 \mathrm{~m} / \mathrm{s}\). Both pieces of fruit hit the ground simultaneously. Calculate the common height from which the fruits were released. Ignore the effects of air resistance.

Short Answer

Expert verified
To calculate the height, use the equation \(h = v_{initial} * t + 0.5 * g * t^2\), and insert the final values for time obtained by solving the quadratic equation. View the solution for each step to check how each part of the problem is managed.

Step by step solution

01

Familiarization

First, familiarize yourself with the key quantities involved. These are the starting speeds, which are \(0 \, \mathrm{m/s}\) for the apple and \(28 \, \mathrm{m/s}\) for the orange, the acceleration due to gravity (\(9.8 \, \mathrm{m/s^2}\)), and the time delay \(1.25 \, \mathrm{s}\).
02

Equate dropping times

Next, we need to understand that both fruits hit the ground at the same time, so the time it took for the apple to hit the ground is the same as the time it took for the orange to hit the ground plus the initial delay of \(1.25 \, \mathrm{s}\). We can write this as \(t_{apple} = t_{orange} + 1.25 \, \mathrm{s}\).
03

Form equations for the distance fallen

We then need to form equations that describe the distance fallen by each fruit. We use the standard kinematics equation \(h = v_{initial} * t + 0.5 * g * t^2\), where \(h\) is the distance fallen, \(v_{initial}\) is the initial speed, \(g\) is the acceleration due to gravity, and \(t\) is the time. For the apple and the orange these are \(h_{apple} = 0.5 * g * t_{apple}^2\) and \(h_{orange} = 28 * t_{orange} + 0.5 * g * t_{orange}^2\) respectively.
04

Set heights equal

As both fruits are thrown from the same height, we now equate the two expressions for height to solve for \(t_{orange}\). Therefore, \(0.5 * g * t_{apple}^2 = 28 * t_{orange} + 0.5 * g * t_{orange}^2\). Substituting from step 2, this equation can be rewritten as \(0.5 * g * (t_{orange} + 1.25 \, \mathrm{s})^2 = 28 * t_{orange} + 0.5 * g * t_{orange}^2\).
05

Solve for time

Solve the previous equation for \(t_{orange}\). This will be a quadratic equation in \(t_{orange}\), which can be solved using standard quadratic equation methods.
06

Calculate height

Once you have found the value of \(t_{orange}\), substitute it back into the equation for \(h_{orange}\) to find the common height.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinematics
Kinematics is a branch of physics and part of classical mechanics. It describes the motion of objects without considering the causes of this motion. It's crucial to understanding projectile motion, where objects are projected into the air.
In kinematics for projectile motion, we often deal with several key quantities:
  • Initial velocity (\( v_{initial} \)): The speed at which an object starts its motion.
  • Time (\( t \)): Duration that the object is in motion.
  • Displacement (\( h \)): The distance the object travels.
  • Acceleration (\( a \)): The rate of change of velocity, with Earth's gravity typically being the only acceleration acting on freely falling objects.

The main kinematic equation for the displacement of an object under constant acceleration used in the exercise is:\[h = v_{initial} \cdot t + \frac{1}{2} \cdot g \cdot t^2\]This equation shows how height, initial velocity, and time work together with acceleration due to gravity to describe projectile motion.
Acceleration due to Gravity
Acceleration due to gravity is a fundamental concept when dealing with objects in free fall near Earth's surface. It is the force that pulls objects downwards and is typically denoted by \( g \).
The standard value of \( g \) on Earth is \( 9.8 \, \mathrm{m/s^2} \). This means that any object falling freely (ignoring air resistance) will accelerate at this rate. Every second that an object is in free fall, its speed will increase by \( 9.8 \, \mathrm{m/s} \).
For projectile motion calculations, \( g \) is crucial, as it allows us to determine how fast an object speeds up as it falls, which impacts how long it takes to hit the ground and how far it travels.
Quadratic Equations
In physics, especially in projectile motion problems, quadratic equations often pop up when you solve for time or other unknowns. A quadratic equation is one of the form:\[a \cdot x^2 + b \cdot x + c = 0\]Where \( a \), \( b \), and \( c \) are constants, and \( x \) represents one of the unknowns.
To solve these, we can use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The "+/-" symbol denotes two potential solutions, which means there can be two different times, velocities, or positions that satisfy the equation depending on the context.
In the exercise, the quadratic equation helps determine the time the orange takes to hit the ground. By setting two expressions for height equal, a quadratic equation is formed, allowing us to solve for \( t_{orange} \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A jet takes off from SFO (San Francisco, CA) and flies to YUL (Montréal, Quebec). The distance between the airports is \(4100 \mathrm{~km}\). After a 1 -h layover, the jet returns to San Francisco. The total time for the round- trip (including the layover) is \(11 \mathrm{~h}, 52 \mathrm{~min}\). If the westbound trip (from YUL to SFO) takes 48 more minutes than the eastbound portion, calculate the time for each leg of the trip. What is the average speed of the overall trip? What is the average speed without the layover?

• Sports In April 1974 , Steve Prefontaine completed a \(10-\mathrm{km}\) race in a time of \(27 \mathrm{~min}, 43.6 \mathrm{~s}\). Suppose "Pre" was at the \(9-\mathrm{km}\) mark at a time of \(25 \mathrm{~min}\) even. If he accelerates for \(60 \mathrm{~s}\) and maintains the increased speed for the duration of the race, calculate the acceleration that he had. Assume his instantaneous speed at the \(9-\mathrm{km}\) mark was the same as his overall average speed at that time.

Biology A black mamba snake has a length of \(4.3\) \(\mathrm{m}\) and a top speed of \(8.9 \mathrm{~m} / \mathrm{s}\) ! Suppose a mongoose and a black mamba find themselves nose to nose. In an effort to escape, the snake accelerates at \(18 \mathrm{~m} / \mathrm{s}^{2}\) from rest. (a) How much time does it take for the snake to reach its top speed? (b) How far does the snake travel in that time? (c) Has the tail of the snake gone past the mongoose or does the mongoose have a chance to catch the black mamba?

What happens to an object's velocity when the object's acceleration is in the opposite direction to the velocity? SSM

•••Calc The acceleration versus time function for an object that starts from rest at \(t=0\) is given by $$ a(t)=6 \mathrm{~m} / \mathrm{s}^{2}+\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t $$ (a) Determine \(v(t)\). (b) Calculate the velocity of the object when \(t=5 \mathrm{~s}\). (c) Determine the displacement of the object from its starting point after \(5 \mathrm{~s}\) has elapsed.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.