# Determine the acceleration of a rocket.

1. Jul 9, 2012

### Squizzel

1. The problem statement, all variables and given/known data
You must determine the acceleration of a rocket so that its equipment can be deisgned to survive. The rocket will have a burn time of t = 30secs, during which time it flies straight up with a constant acceleration a; call this "Phase 1" of the experiment. After the fuel is exhusted, the rocket will enter free-fall; call this "Phase 2". The total time of flight of the rocket must be T = 300secs.

A) What should you make the acceleration of the rocket, a, when the engine is on?
B) What is the maximum altitude of the rocket.

2. Relevant equations

x=xo + vot + 1/2 at^2

3. The attempt at a solution

I just started, but so far I am setting up the phase 1 variable chart. I am confused by the straight up part, does this mean that the change in x is 0? that would give me 3 0s on the x side of the chart.

2. Jul 9, 2012

### Squizzel

So I got the initial V for the y using x=xo + vot + 1/2 at^2, and I got 294.3.

3. Jul 10, 2012

### ehild

Hi Squizzel,

How did you get that value for the initial V?

ehild

4. Jul 10, 2012

### Squizzel

I used v = vo + at , the final velocity will be zero as it will enter free fall.

edit : Oh, and Hi!

5. Jul 10, 2012

### ehild

What is t?

ehild

6. Jul 10, 2012

### Squizzel

30 s, trying to figure out the phase one velocity.

7. Jul 10, 2012

### ehild

The rocket engine works for 30 s. During that time the rocket accelerates upward with an unknown acceleration, , from zero velocity to some v1, reaching height y1=h. This is the first stage. From here on, the rocket moves like a stone projected vertically up with initial velocity v0=v1, and initial height y0=h for 270 s, when it hits the ground (y=0)

Write up the equations for velocity v1 and height y1 for the first stage with t=30.
Use the expressions for h and v0 for the second stage.

ehild

8. Jul 10, 2012

### Squizzel

There are two different equations for v1 correct? One in the x and y.

9. Jul 10, 2012

### ehild

There are two different equation for the two stages. You can denote the displacement by x and y, but do not forget that both are vertical.

ehild

10. Jul 10, 2012

### Squizzel

For the first equation is there any change in X? I am confused because I thought there should be two equations for each stage. A x and y one for both.

11. Jul 10, 2012

### ehild

If you consider x as the horizontal displacement, forget it. The motion is vertical, there is no displacement horizontally.

ehild

12. Jul 10, 2012

### Squizzel

ok so I am applying this equation for y=yo + vot + 1/2 at^2 , now I am trying to find alceration in the x direction right? because the acceleration in the y direction is always -9.81 I thought.

13. Jul 10, 2012

### ehild

There is no x direction. The rocket moves vertically. You need to find the acceleration in the first stage.
During the second stage the acceleration is -9.81.

ehild

14. Jul 10, 2012

### Squizzel

First the first stage, shouldn't I take into account the -9.81 effect of gravity?

15. Jul 10, 2012

### Squizzel

For V1, I get V= Vo + At

V = (a-g)t

16. Jul 10, 2012

### ehild

I do not understand. Do you mean the first stage? Why do you use g?

ehild

17. Jul 10, 2012

### Squizzel

Yes, I thought that g was always in affect when you look at acceleration in the y direction.

This is what i worked out, does this make sense:

part a
v=0 at top point
v= vo+(a-g)t
vo= (a-g)*30 ..........P
v^2 = u^2 +(a-g)s
vo^2 = (a-g)s .............Q
so v at ground= 9.8*270 = 2646m/s
v^2 = 2gs
s=2646^2/(2*9.8) = 357210 m
from Q
vo^2 = (a-g)357210
from P
vo^2 = (a-g)^2*900
so (a-g)357210 = (a-g)^2*900
a-g = 396.9
so a= 406.71m/s^2

Last edited: Jul 10, 2012
18. Jul 10, 2012

### azizlwl

1. Total displacement is zero.
2. Rocket travelled with 2 constant acceleration +a and -g

19. Jul 10, 2012

### ehild

If you like you can call the upward acceleration "a-g". Then "a" would be the acceleration of the rocket in free space. But the problem says that the rocket flyes upward with constant acceleration a.

The rocket starts from zero velocity. Have you seen the launch of a rocket? What is your vo then?

The end of the first stage is when the fuel burns out. It does not mean that the rocket reaches the top height then: it will rise till the velocity becomes zero, but its acceleration is -9.81 m/s^2 during this time, and this rising motion also belongs to the second stage.

ehild

20. Jul 10, 2012

### Squizzel

OK so my first equation would v/30 = a because V = Vo + at and Vo = 0.