/*! 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 40 A lunar lander is making its des... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 40

A lunar lander is making its descent to Moon Base I (\(\textbf{Fig. E2.40}\)). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 m above the surface and has a downward speed of 0.8 m/s.With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 m/s\(^{2}\).

Short Answer

Expert verified
The speed of the lander just before touching the surface is approximately 4.08 m/s.

Step by step solution

01

Identify Known Values

First, let's identify the given values in the problem. The initial downward speed of the lander is 0.8 m/s (denoted as \( v_i\)). It descends from a height of 5.0 m above the surface (distance \( d = 5.0 \text{ m} \)). The gravitational acceleration on the moon is 1.6 m/s\(^2\).
02

Choose an Appropriate Equation

To find the final speed just before the lander touches the surface, we will use the kinematic equation:\[ v_f^2 = v_i^2 + 2ad \]where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the displacement.
03

Substitute Known Values

Substitute the known values into the kinematic equation: \( v_i = 0.8 \text{ m/s} \), \( a = 1.6 \text{ m/s}^2 \), and \( d = 5.0 \text{ m} \).Plugging them into the equation:\[ v_f^2 = (0.8)^2 + 2 \times 1.6 \times 5.0 \]
04

Calculate

Calculate the expression from Step 3: First calculate \( (0.8)^2 \): \( (0.8)^2 = 0.64 \)Then calculate \( 2 \times 1.6 \times 5.0 \):\( 2 \times 1.6 \times 5.0 = 16.0 \).Add these together:\[ v_f^2 = 0.64 + 16.0 = 16.64 \]
05

Solve for Final Velocity

To find \( v_f \), take the square root of \( v_f^2 \):\[ v_f = \sqrt{16.64} \approx 4.08 \text{ m/s} \]
06

Check Units and Answer

Ensure the calculated speed is in meters per second, which it is. The final speed of the lander just before it touches the surface of the moon is approximately 4.08 m/s.

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.

Lunar Descent Dynamics
Lunar descent dynamics involves understanding how a spacecraft, such as a lunar lander, behaves as it approaches the lunar surface. This process includes modifications in speed and direction due to gravitational forces and engine thrust. During the landing phase:
  • The lander often uses retro-thrust engines to slow down during its descent.
  • Once the engines are cut off, the lander experiences free fall under the influence of lunar gravity.
The challenge is to manage these transitions smoothly to ensure a safe landing. Understanding the descent dynamics is crucial for ensuring the spacecraft lands at a manageable velocity to prevent damage or accidents.
Gravitational Acceleration on the Moon
Gravitational acceleration on the moon is significantly weaker than on Earth. It is a crucial factor in determining how objects will fall and the speed they will reach during free fall. The moon's gravity:
  • Is approximately 1/6th of Earth's gravitational pull, measured at about 1.6 m/s虏.
  • Affects the trajectory and speed of any object in free fall, such as the lunar lander in our scenario.
This lighter gravitational pull means that objects fall slower on the moon than they would on Earth. In a practical scenario like our lunar descent, it is this gravitational constant that dictates how quickly the lander speeds up once engine thrust is cut off and it begins its free fall to the lunar surface.
Kinematic Equations
Kinematic equations are a set of mathematical expressions used to describe motion. These equations are essential in predicting the future positions and velocities of moving objects under constant acceleration, like our lunar lander.The primary equation used here is:\[v_f^2 = v_i^2 + 2ad\]Where:
  • \(v_f\) is the final velocity we want to determine.
  • \(v_i\) is the initial velocity, given as 0.8 m/s in the problem.
  • \(a\) stands for acceleration, specifically the moon's gravitational acceleration of 1.6 m/s虏 in this scenario.
  • \(d\) denotes displacement, which is the height of 5.0 m the lander falls.
By substituting the known values into this equation, we efficiently calculate the final speed of the lander when it reaches the lunar surface. Understanding how to use kinematic equations is essential for analyzing any movement involving uniform acceleration, such as our lunar lander's free-fall scenario.

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 student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s\(^2\). (a) For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x =\) 0 at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 m/s, will she catch the bus? (f) What is the \(minimum\) speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

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 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.

Starting from the front door of a ranch house, you walk 60.0 m due east to a windmill, turn around, and then slowly walk 40.0 m west to a bench, where you sit and watch the sunrise. It takes you 28.0 s to walk from the house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your (a) average velocity and (b) average speed?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. Ignore air resistance. (a) If the height of the building is 20.0 m, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_0\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_0\) is 6.0 m/s and (ii) \(v_0\) is 9.5 m/s? (c) If \(v_0\) is greater than some value \(v_{max}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{max}\). The value \(v_{max}\) has a simple physical interpretation. What is it? (d) If \(v_0\) is less than some value \(v_{min}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{min}\). The value \(v_{min}\) also has a simple physical interpretation. What is it?

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Oscillations

Read Explanation

Rotational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Electricity and Magnetism

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.