How High Does a Two-Stage Rocket Reach?

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SUMMARY

The discussion focuses on calculating the maximum height and time of descent for a two-stage rocket. The first stage accelerates at 3.50 m/s² for 25 seconds, reaching an initial velocity of 132.5 m/s for the second stage. The maximum height reached after the second stage is calculated using the equation X = 895.7 m. The discussion emphasizes the need to solve the problem in two parts: first for the ascent during stage one and then for the ascent and descent during stage two, incorporating gravitational effects.

PREREQUISITES
  • Understanding of kinematic equations for uniformly accelerated motion
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Familiarity with the concept of two-stage rocket dynamics
  • Ability to extract and organize information from word problems
NEXT STEPS
  • Study kinematic equations in detail, especially V² = Vo² + 2a(X - Xo)
  • Learn about projectile motion and its applications in rocketry
  • Explore the effects of gravity on objects in free fall
  • Practice solving multi-stage motion problems in physics
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Students studying physics, particularly those focusing on kinematics and dynamics, as well as educators seeking to enhance their teaching of motion concepts in rocketry.

eeriana
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Homework Statement


In the first stage of a two stage rocket, the rocket is fired fro, the launch pad starting from rest but with a constant acceleration of 3.50 m/s^2 upward. At 25.0 s after launch, the rocket fires the second stage, which suddenly boosts its speed to 132.5 m/s upward. This firing uses up all the fuel, however, so then the only force acting on the rocket is gravity. negligible air resistance. a) find the maximum height that the stage-two rocket reaches above the launch pad. b) how much time after the stage-two firing will it take for the rocket to fall back to the launch pad? c) how fast with the stage two rocket be moving just as it reaches the launch pad.


Homework Equations


These are the equations I have:
V^2=Vo^2 +2a(X-Xo)
V=Vo+at
X=Xo+volt+1/2at^2
X-Xo=1/2(Vo+V)t


The Attempt at a Solution



Hi, I am new here, and am word problem challenged! I have a hard time extracting the information and setting it up.

I have Xo=0 a=3.5m/s^2 t=25s Vo=0
stage two Vo=132 m/s V=0 Xo=0 a=-9.8m/s^2 x=?

0=(132.5m/s)^2 + 2(-9.8m/s^2)(X-0)
0=17556.25 m^2/s^2 -19.6m/s^2x
X= 895.7m

the I get
t=? Xo= 895.7m X=0 Vo= 132.5 m/s V=0


Sorry I don't have more, but that is where I get stuck... The second stage is throwing me. I don't know if I am extracting the information correctly and then getting it set up.

Any help or suggestions to point me in the right direction would be appreciated.

Thanks...
Amy
 
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The key is to solve two problems - one for stage 1 and the other for stage 2.

Stage 1: Xo=0 a=3.5m/s^2 t=25s Vo=0 correct, V(t=25s) = 132.5 m/s

Stage 2: Vo=132.5 m/s a=-9.8m/s^2 x=? correct, but V=0 Xo=0 are not correct.

The rocket is at some height after accelerating with stage 1 for 25s. One needs to solve for that height for the stage 1 operation.

Stage 2 has basically starts at the height (from part 1), with an initial velocity of 132.5 m/s and gradually decelerates to maximum altitude.

So for part 1 and part 2, write the appropriate equations.

See this for some reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
 
Last edited:

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