How high and how fast will a 25 kg rocket fly without air resistance?

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Homework Help Overview

The problem involves a 25 kg rocket generating 400 N of thrust, designed to burn fuel for 40 seconds, and seeks to determine its final vertical velocity, maximum height, and time until it reaches the ground, while neglecting air resistance and the mass of the fuel.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss summing forces to find acceleration and use kinematic equations to calculate final velocity and height. Questions arise regarding the inclusion of gravitational acceleration and the interpretation of flight time.

Discussion Status

Some participants have offered guidance on the calculations, while others are exploring the implications of including gravitational effects and clarifying the time of flight. There is a recognition of confusion regarding the time until the rocket reaches the ground after fuel burnout.

Contextual Notes

Participants note the assumption of neglecting air resistance and the mass of the fuel, which may influence the calculations and interpretations of the rocket's motion.

David Donald
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Homework Statement


A 25 kg Rocket generates 400 N of thrust. It is designed to burn its fuel for
40 seconds. If launched vertically what will its final vertical velocity be when the fuel runs out? How high can it fly? How long until it reaches the ground? Neglect mass of fuel, and air resistance.

Please check over my work I feel like I may be making a mistake somewhere

Homework Equations


Kinematics Equations

The Attempt at a Solution


First I summed the forces in the Y direction
ΣFy = Ft - Fg = ma
= (400N - 245N)/(25kg) = a = 6.2 m/s^2

Final Vertical Velocity
Vfy = Voy + at
Vfy = 6.2 m/s^2 * 40s = 248 m/s

(Height when fuel runs out)

Yf = Yo + Voy*t + (1/2)a t^2
Yf = (1/2)*(6.2 m/s^2) (40)^2 = 4960 meters

(FINAL HEIGHT)

Vfy = Voy + at
0 = 248 m/s - 9.8m/s^2 t = 25.31 seconds <---- Time of Final Height

Yf = Yo + Voy*t + (1/2)at^2
Yf = 4960m + 248 m/s * 25.31 s = 11385.7 meters

(How Long until it reaches the ground)
I'm confused how to solve this part...
 
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Everything looks good up to here:
David Donald said:
Yf = Yo + Voy*t + (1/2)at^2
Yf = 4960m + 248 m/s * 25.31 s = 11385.7 meters
What about the (1/2)at2 term in the equation above?

(How Long until it reaches the ground)
I'm confused how to solve this part...
Can you specify in more detail what is confusing your here?
 
I forgot to include the negative acceleration due to gravity right?
i'm not sure how to find the time of max flight...
 
David Donald said:
I forgot to include the negative acceleration due to gravity right?
Right.
i'm not sure how to find the time of max flight...
Not sure what you mean by "max flight". Try to find the time from "burn out" until the rocket hits the ground. During this time the rocket is in free fall.

{EDIT: Or, you can try to find the time from the point of max height until the rocket hits the ground. This will be easier. You already know the time to reach max height.}
 
Nevermind I figured it out, thank you!
 

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