Determine the acceleration of a rocket.

In summary: the y value at the time of landing is y=0 because the rocket just land and there is no displacement horizontally.
  • #1
Squizzel
29
0

Homework Statement


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.

Homework Equations



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

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.
 
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  • #2
So I got the initial V for the y using x=xo + vot + 1/2 at^2, and I got 294.3.
 
  • #3
Hi Squizzel,

How did you get that value for the initial V? ehild
 
  • #4
I used v = vo + at , the final velocity will be zero as it will enter free fall.

edit : Oh, and Hi!
 
  • #5
What is t?

ehild
 
  • #6
30 s, trying to figure out the phase one velocity.
 
  • #7
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
There are two different equations for v1 correct? One in the x and y.
 
  • #9
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
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
If you consider x as the horizontal displacement, forget it. The motion is vertical, there is no displacement horizontally.

ehild
 
  • #12
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
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
First the first stage, shouldn't I take into account the -9.81 effect of gravity?
 
  • #15
For V1, I get V= Vo + At

V = (a-g)t
 
  • #16
I do not understand. Do you mean the first stage? Why do you use g?

ehild
 
  • #17
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
time taken for free fall = 300-30= 270s
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:
  • #18
1. Total displacement is zero.
2. Rocket traveled with 2 constant acceleration +a and -g
 
  • #19
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
OK so my first equation would v/30 = a because V = Vo + at and Vo = 0.
 
  • #21
What is v/30?

ehild
 
  • #22
V/t = a for the first phase t is 30.
 
  • #23
So the acceleration during the first phase is a, and the velocity after 30 seconds is v=30a. What is the hight of the rocket at that instant?

ehild
 
  • #24
Y = 1/2 at,^²
 
  • #25
To find the V for the first equation, I know that it is the initial velocity in the second equation.

so for the 2nd equation I got V = Vo + at

0 = Vo + (-9.81)(270) v = 0 because that is when the rocket lands. with this equation I got that Vo = 2648.7
 
  • #26
Plugging that back into my first equation I get that V = a(30) and a = 88.3
 
  • #27
what is did was incorrect
sorry
 
Last edited:
  • #28
Squizzel said:
To find the V for the first equation, I know that it is the initial velocity in the second equation.

so for the 2nd equation I got V = Vo + at

0 = Vo + (-9.81)(270) v = 0 because that is when the rocket lands. with this equation I got that Vo = 2648.7

The velocity is not zero when the rocket lands.

For the second stage, vo = 30a and yo =302a/2=450a.
The second stage is under the effect of gravity alone, "free fall" but the initial velocity is upward,so the rocket will rise with decreasing velocity. After the velocity becomes zero, the rocket turns back and begins falling down.
The displacement is y=yo+vot-9.81/2 t2. If t=270s, the rocket reaches the ground, y=0. Substituting vo=30a and yo=450a, t=270, you get an equation for a.

ehild
 
  • #29
srry incorrect wrking was put here
 
Last edited:
  • #30
roovid said:
i looked at final stage first..the free fall
assuming the rocket fell exactly aft the fuel finished from question

total flight = 300s
time to fall during free flight = 300-30
=270

The rocket will not turn back immediately as the fuel burns out. It will rise till the velocity becomes zero. So the real time of fall is less than 270 s.
"free fall" means motion under the gravity of Earth alone.

ehild
 
  • #31
oh my bad then
i thought for calculation purposes only they wanted u to assume tohe rocket fell aft fuel ran out
thanks for clearing up the free fall term for me
 
  • #32
roovid said:
oh my bad then
i thought for calculation purposes only they wanted u to assume tohe rocket fell aft fuel ran out
thanks for clearing up the free fall term for me

How can a rocket change its velocity to zero in no time, without fuel? :biggrin: Even the problem maker can not assume nonsense...

ehild
 
  • #33
lol i kno
but some of the question i am doing are makin me assume some weird stuff
for basic calculation purposes
sorry.my error
thnx fr the correction again
 
  • #34
You are welcome. :smile:

ehild
 
  • #35
So a = 176 ms^2?
 

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