How High Does a Two-Stage Rocket Reach?

[eeriana]

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

 

[Astronuc]

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

 

[m2003]

Hi Amy,

First of all, great job on setting up the equations correctly! Your approach is correct so far.

To find the maximum height the stage-two rocket reaches, you can use the equation for displacement: X=Xo+Vot+1/2at^2. In this case, Xo=895.7m, Vo=132.5m/s, and a=-9.8m/s^2. The only unknown is t, so you can solve for it by plugging in the given values and solving for t. This will give you the time at which the rocket reaches its maximum height.

To find the time it takes for the rocket to fall back to the launch pad, you can use the same equation but with X=0 (since the rocket falls back to the launch pad, its final displacement is 0). You already know the value for Xo, Vo, and a, so you can solve for t and this will give you the time it takes for the rocket to fall back to the launch pad.

To find the final velocity of the rocket just as it reaches the launch pad, you can use the equation V=Vo+at, where Vo=132.5m/s, a=-9.8m/s^2, and t is the time you found in the previous step. This will give you the final velocity of the rocket just as it reaches the launch pad.

I hope this helps! Let me know if you have any other questions. Good luck with your problem solving!