How High Does a Two-Stage Rocket Reach?

AI Thread Summary
A two-stage rocket accelerates at 3.5 m/s² for 25 seconds before firing its second stage, reaching an initial speed of 132.5 m/s. To find the maximum height, calculations must first determine the height achieved during the first stage, using the appropriate kinematic equations. The second stage then involves calculating the additional height gained before the rocket decelerates to a stop due to gravity. Finally, the time for the rocket to fall back to the launch pad and its speed upon landing can be determined. Properly setting up the equations for both stages is crucial for solving the problem effectively.
eeriana
Messages
15
Reaction score
0

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+Vot+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
 
Physics news on Phys.org
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top