Two Stage Rocket: Max Height Above Launch Pad

  • Thread starter Thread starter Toranc3
  • Start date Start date
  • Tags Tags
    Rocket
Click For Summary

Homework Help Overview

The problem involves a two-stage rocket launched from a pad, with specific accelerations and time intervals for each stage. The goal is to determine the maximum height reached above the launch pad after both stages of the rocket's flight.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for the first and second stages of the rocket's ascent, including the use of kinematic equations. Some suggest alternative methods for calculating height using average velocity.

Discussion Status

There is acknowledgment of the calculations presented, with some participants confirming the results while also suggesting that the rocket continues to rise after the second stage finishes firing. Further parts of the question are anticipated to be addressed later.

Contextual Notes

Participants note that air resistance is negligible and that the only force acting on the rocket after the second stage is gravity. There is an indication that the problem may have additional components that have not yet been discussed.

Toranc3
Messages
189
Reaction score
0

Homework Statement



In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s^2 upward. At 25.0 seconds after launch, the second stage fires for 10 seconds, which boosts the rocket's velocity to 132.5m/s upward at 35 seconds after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance is negligible.

A) Find the maximum height that the stage two rocket reaches above the launch pad.

Homework Equations



y=yo + vo*t +1/2*a*t^(2)

vy=vo + a*t




The Attempt at a Solution



First stage:
y=yo + vo*t +1/2*a*t^(2)
y=1/2(3.5m/s^(2))*(25s)^(2)
y=1093.75m

Vy=Vo + a*t
Vy=3.50m/s^(2)*25s= 87.5m/s
Final velocity of first stage=87.5m/s

Second Stage:
Initial velocity of second stage=87.5m/s
vy=vo+a*t
132.5m/s=87.5m/s +a*10sec
a=4.5m/s^(2)

New acceleration=4.5m/s^(2)

y=yo + vo*t +1/2*a*t^(2)

y=1093.75m + 87.5m/s*10s + 1/2(4.5m/s^(2))*100s= 2193.75m

Is this the correct answer? Thanks!
 
Physics news on Phys.org
Yes, it looks right. It would have been slightly simpler to compute the stage two height delta by average velocity: 2s = (v0+vf)t
I assume there are more parts to the question.
 
haruspex said:
Yes, it looks right. It would have been slightly simpler to compute the stage two height delta by average velocity: 2s = (v0+vf)t
I assume there are more parts to the question.

Thanks and yes there are more parts to this question. I am going to work on them and then post them back on here.
 
2193.75m is the height after the second stage finished firing - but the rocket still has a velocity upwards, and will continue to increase its height afterwards.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
11
Views
2K
Replies
6
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
23
Views
2K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K