/*! 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 During your summer internship fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 78

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of \(960 \mathrm{~m}\) above the earth's surface. The rocket's engines give the rocket an upward acceleration of \(16.0 \mathrm{~m} / \mathrm{s}^{2}\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

Short Answer

Expert verified
Solving the equation \(t = \sqrt{\frac{2d}{a}}\) with \(d=960\) m and \(a=16.0\) m/s虏 gives \(t = 30\) s. Thus, the rocket's engines need to fire for 30 seconds to reach the required altitude.

Step by step solution

01

Understand and organize the given information

The rocket is launched from rest so initial velocity \(v_i = 0\) . The height it is supposed to reach is \(d = 960\) meters. The acceleration during the rocket's ascend is \(a = 16.0 \, m/s^2\) . We need to find the time \(T\) under these conditions.
02

Apply the kinematic equation

The kinematic equation for distance under constant acceleration is: \(d = v_i t + 0.5 a t^2\). Given that the initial velocity \(v_i = 0\) , the equation simplifies to: \(d = 0.5 a t^2\) . We need to solve this for \(t\).
03

Solve for T

Rearrange the simplified equation to find \(t\): it gives \(t^2 = \frac{2d}{a}\). Taking the square root of both sides gives \(t = \sqrt{\frac{2d}{a}}\). Substitute the known values for \(d\) (960 m) and \(a\) (16.0 m/s虏), which allows you to compute \(t\).

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 91影视!

Key Concepts

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

Rocket Motion
Launch a rocket involves several physics principles. When a rocket takes off, it starts from rest. This means its initial velocity (\(v_i\) is zero. The rocket's engines create a force, resulting in an upward acceleration. In this problem, the given acceleration is \(16.0\, \mathrm{m/s}^2\).
The distance the rocket should travel upward is specified as \(960\, \mathrm{m}\). The goal is to calculate the time \(T\) for which the engines need to apply this force for the rocket to reach that height.
The calculation assumes no air resistance, which simplifies the problem. The main hurdle is determining how long the engines need to produce this constant thrust. By understanding the concepts of motion and forces, a rocket designer can calculate the necessary parameters for a successful launch.
Free Fall
After the rocket's engines shut off, the rocket enters a new phase of motion known as free fall. In free fall, the only force acting on the rocket is gravity. This means the acceleration due to gravity, often denoted by \(g\), is in effect, with a constant value of \(9.8\, \mathrm{m/s}^2\) directed downwards towards the earth.
During this phase, the upward motion gradually slows down until the rocket momentarily stops at its peak, before accelerating back downwards.
  • The rocket's velocity at the peak height is zero.
  • The distance covered during the engine's thrust is the point at which free fall begins.
In practical terms, understanding and calculating the free fall phase is crucial for predicting the rocket's trajectory and ensuring it returns safely after reaching its desired altitude.
Constant Acceleration
Constant acceleration is a key concept in understanding the motion of objects, including rockets. When the rocket is accelerating upwards with a constant value of \(16.0\, \mathrm{m/s}^2\), it means that the velocity of the rocket increases steadily by \(16.0\, \mathrm{m/s}\) every second.
When using the simplified kinematic equation \(d = 0.5 a t^2\),
  • \(d\) represents the distance traveled.
  • \(a\) is the constant acceleration.
  • \(t\) is the time duration for which the acceleration is applied.
Rearranging this equation to solve for time aids in determining how long the engines need to run to reach a specific altitude. This calculation is crucial for effective flight planning and ensuring that the rocket reaches its intended height without running out of fuel or power prematurely.

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 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 \(7.0 \mathrm{~s}\) 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?

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g}\) ?

A \(7500 \mathrm{~kg}\) rocket blasts off vertically from the launch pad with a constant upward acceleration of \(2.25 \mathrm{~m} / \mathrm{s}^{2}\) and feels no appreciable air resistance. When it has reached a height of \(525 \mathrm{~m}\), its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

A block moving on a horizontal surface is at \(x=0\) when earth's surface. At \(1.15 \mathrm{~s}\) after liftoff, the rocket clears the top of its launch platform, \(63 \mathrm{~m}\) above the \(\mathrm{ground.}\) After an additional \(4.75 \mathrm{~s},\) it is \(1.00 \mathrm{~km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 s part of its flight and (b) the first \(5.90 \mathrm{~s}\) of its flight.

A tennis ball on Mars, where the acceleration due to gravity is \(0.379 g\) and air resistance is negligible, is hit directly upward and returns to the same level \(8.5 \mathrm{~s}\) later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball's vertical position, vertical velocity, and vertical acceleration as functions of time while it's in the Martian air.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Wave Optics

Read Explanation

Engineering Physics

Read Explanation

Medical Physics

Read Explanation

Turning Points in Physics

Read Explanation

Physical Quantities And Units

Read Explanation
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.