/*! 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 46 An egg is thrown nearly vertical... [FREE SOLUTION] | 91Ó°ÊÓ

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An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0 \(\mathrm{m}\) below its starting point 5.00 s after it leaves the thrower's hand. Air resistance may be ignored. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.

Short Answer

Expert verified
Initial speed: 14.53 m/s; Max height: 10.76 m; Velocity at peak: 0 m/s; Acceleration at peak: 9.81 m/s² (downward).

Step by step solution

01

Identify Known Values

To solve the problem, start by noting the known values. The egg falls 50.0 m below the starting point over 5.00 s. Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \), acts downward. These values set up the framework for the kinematic equations.
02

Apply the Kinematic Equation for Distance

Use the kinematic equation \( y = v_{0}t + \frac{1}{2}gt^2 \) to find the initial velocity, where \( y = -50.0 \text{m} \) (as it's below the starting point), and \( t = 5.00 \text{s} \). This becomes \( -50.0 = v_{0} \times 5.00 + \frac{1}{2} (-9.81) \times (5.00)^2 \).
03

Solve for Initial Velocity

Rearrange the equation: \( -50.0 = 5v_{0} - 122.625 \). Solving for \( v_{0} \), we get \( v_{0} = \frac{72.625}{5} \approx 14.53 \, \text{m/s} \). This is the initial speed of the egg.
04

Calculate Maximum Height

At the highest point, the velocity is 0. Using the equation \( v^2 = v_{0}^2 + 2g(y - y_0) \) and knowing that \( v = 0 \) at the peak, solve for \( y - y_0 \). Thus, \( 0 = (14.53)^2 - 2 \times 9.81 \times (y - 0) \) gives \( y = \frac{(14.53)^2}{2 \times 9.81} \approx 10.76 \, \text{m} \).
05

Determine Velocity at the Highest Point

The velocity at the highest point is 0 \( \text{m/s} \) since it's the point of change in direction from up to down.
06

Find Acceleration at the Highest Point

The acceleration due to gravity is constant throughout the motion. Therefore, its magnitude at the highest point is \( 9.81 \, \text{m/s}^2 \). The direction is downward.
07

Sketch Graphs

Illustrate the motion with three graphs: 1. \( a_y-t \): A horizontal line at \( -9.81 \, \text{m/s}^2 \).2. \( v_y-t \): A line starting at 14.53 \( \text{m/s} \), decreasing linearly to 0 \( \text{m/s} \) in half the total time, then continuing into negative values.3. \( y-t \): A parabolic shape, peaking at the maximum height of 10.76 \( \text{m} \) and then descending.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental in describing the motion of objects. When dealing with objects in free fall or projectile motion, these equations help determine variables like displacement, velocity, and time. They incorporate both the initial conditions of the motion and the effects of acceleration. The key equations are:
  • Displacement: \[ y = v_{0}t + \frac{1}{2}gt^2 \]
  • Final velocity: \[ v = v_{0} + gt \]
  • Final velocity squared: \[ v^2 = v_{0}^2 + 2g(y - y_0) \]
In these equations, \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity, \( y \) is the displacement, and \( t \) is the time elapsed. Understanding how to choose and rearrange these equations allows us to solve problems like finding the initial velocity of a projectile.
Acceleration due to Gravity
Acceleration due to gravity is a constant that determines the rate at which an object accelerates towards the Earth. It is denoted as \( g \) and is approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface. This uniform acceleration applies to all objects in free fall near the Earth, regardless of their mass. Hence, in problems involving projectile motion, such as the egg tossed from a building, this constant determines the motion of the object throughout.
One crucial point is that this acceleration is always directed downwards, meaning that if we assign upward directions as positive, \( g \) would be negative. Understanding the role of this constant helps us predict how the object's velocity changes over time and affects its path through kinematic equations.
Initial Velocity Calculation
Calculating the initial velocity of an object is an important step in analyzing motion. For our egg projectile problem, the initial velocity \( v_0 \) can be determined using the kinematic equation for displacement:
  • \[ y = v_{0}t + \frac{1}{2}gt^2 \]
Here, \( y \) is the final position relative to the starting point, and \( t \) is the time taken to reach that position. By rearranging this equation, we solve for \( v_0 \), the initial speed.
In our exercise, after substituting known values and solving, we found \( v_0 \approx 14.53 \text{ m/s} \). This initial velocity indicates the speed at which the egg had to be thrown upwards initially to travel to the specified point in the stated time. Accurately determining \( v_0 \) is crucial in predicting other aspects of the object’s motion, like maximum height and time of flight.
Graphical Analysis of Motion
Graphical analysis of motion provides a visual understanding of the dynamics involved in projectile motion. It includes plotting various graphs — such as acceleration vs. time (\( a_y-t \)), velocity vs. time (\( v_y-t \)), and position vs. time (\( y-t \)).
  • For the acceleration graph (\( a_y-t \)), you would see a constant line at \( -9.81 \, \text{m/s}^2 \) indicating uniform acceleration due to gravity.
  • In the velocity graph (\( v_y-t \)), the line starts at the initial velocity of \( 14.53 \, \text{m/s} \) and linearly decreases to zero (at the peak) before becoming negative as the egg descends.
  • The position graph (\( y-t \)) takes a parabolic shape, rising to a peak (maximum height) before descending back down past the starting point.
Such graphical representations help in better understanding how variables like velocity and acceleration change over time, providing insights beyond numerical analysis. Recognizing these patterns in graphs is fundamental in mastering concepts in projectile motion.

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Most popular questions from this chapter

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2} .\) Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 \(\mathrm{s}\) , Powers shuts off the engine and steps out of the helicopter Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

A painter is standing on scaffolding that is raised at constant speed. As he travels upward, he accidentally nudges a paint can off the scaffolding and it falls 15.0 \(\mathrm{m}\) to the ground. You are watching, and measure with your stopwatch that it takes 3.25 \(\mathrm{s}\) for the can to reach the ground. Ignore air resistance. (a) What is the speed of the can just before it hits the ground? (b) Another painter is standing on a ledge, with hands 4.00 \(\mathrm{m}\) above the can when it falls off. He has lightning-fast reflexes and if the can passes in front of him, he can catch it. Does he get the chance?

A ball is thrown straight up from the ground with speed \(v_{0}\) . At the same instant, a second ball is dropped from rest from a height \(H\) , directly above the point where the first ball was thrown upward. There is no air resistance, (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) so that at the instant when the balls collide, the first ball is at the highest point of its motion.

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

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