Skip to main content
Ch 02: Kinematics in One Dimension
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 2, Problem 56a

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s² for 30 s, then runs out of fuel. Ignore any air resistance effects. What is the rocket's maximum altitude?

Verified step by step guidance
1
Step 1: Calculate the velocity of the rocket at the end of the fuel burn phase using the formula for final velocity under constant acceleration: v=u+at, where u is the initial velocity (0 m/s), a is the acceleration (30 m/s²), and t is the time (30 s).
Step 2: Calculate the altitude reached during the fuel burn phase using the kinematic equation: s=ut+12at², where u is the initial velocity (0 m/s), a is the acceleration (30 m/s²), and t is the time (30 s).
Step 3: Determine the rocket's motion after the fuel runs out. The rocket will continue to ascend due to its velocity at the end of the fuel burn phase. Use the kinematic equation s=v2/g to calculate the additional altitude gained during this phase, where v is the velocity at the end of the fuel burn phase and g is the acceleration due to gravity (9.8 m/s²).
Step 4: Add the altitude from the fuel burn phase to the altitude gained during the coasting phase to find the total maximum altitude of the rocket.
Step 5: Ensure all units are consistent throughout the calculations (e.g., meters, seconds) and verify the results conceptually to confirm the rocket's maximum altitude.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for understanding how the rocket accelerates upward when the engines are firing, as the net force is the difference between the thrust produced by the rocket and the gravitational force acting on it.
Recommended video:
Guided course
06:54
Intro to Forces & Newton's Second Law

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. In this scenario, they can be used to calculate the rocket's velocity and displacement during the powered ascent phase, as well as the subsequent free-fall phase after the fuel runs out, allowing us to determine the maximum altitude reached.
Recommended video:
Guided course
08:25
Kinematics Equations

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. As the rocket ascends, it converts kinetic energy into gravitational potential energy. Understanding this concept is essential for calculating the maximum altitude, as it involves determining the point where the rocket's kinetic energy is fully converted into potential energy before it starts descending.
Recommended video:
Guided course
06:35
Gravitational Potential Energy
Related Practice
Textbook Question

A cheetah spots a Thomson's gazelle, its preferred prey, and leaps into action, quickly accelerating to its top speed of 30 m/s, the highest of any land animal. However, a cheetah can maintain this extreme speed for only 15 s before having to let up. The cheetah is 170 m from the gazelle as it reaches top speed, and the gazelle sees the cheetah at just this instant. With negligible reaction time, the gazelle heads directly away from the cheetah, accelerating at 4.6 m/s² for 5.0 s, then running at constant speed. Does the gazelle escape? If so, by what distance is the gazelle in front when the cheetah gives up?

1
views
Textbook Question

Find an expression for the minimum stopping distance dstop of a car traveling at speed v0 if the driver's reaction time is Treact and the magnitude of the acceleration during maximum braking is a constant abrake.

2
views
Textbook Question

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s². You instantly start running toward the still-open door at 4.5 m/s. What is the maximum time you can wait before starting to run and still catch the bus?

2
views
Textbook Question

A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2. How long does it take to make the complete trip from bottom to top?

Textbook Question

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake?

2
views
Textbook Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 16 s?

1
views