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Determine the acceleration of a rocket.

 
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Jul10-12, 02:18 AM   #18
 

Determine the acceleration of a rocket.


1. Total displacement is zero.
2. Rocket travelled with 2 constant acceleration +a and -g
Jul10-12, 02:44 AM   #19

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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
Jul10-12, 02:58 AM   #20
 
OK so my first equation would v/30 = a because V = Vo + at and Vo = 0.
Jul10-12, 03:44 AM   #21

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What is v/30?

ehild
Jul10-12, 11:21 AM   #22
 
V/t = a for the first phase t is 30.
Jul10-12, 12:07 PM   #23

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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
Jul10-12, 01:01 PM   #24
 
Y = 1/2 at,^²
Jul10-12, 02:11 PM   #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
Jul10-12, 02:15 PM   #26
 
Plugging that back into my first equation I get that V = a(30) and a = 88.3
Jul10-12, 03:28 PM   #27
 
what is did was incorrect
sorry
Jul10-12, 04:03 PM   #28

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Quote by Squizzel View Post
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
Jul10-12, 04:08 PM   #29
 
srry incorrect wrking was put here
Jul10-12, 04:08 PM   #30

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Quote by roovid View Post
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
Jul10-12, 04:10 PM   #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
Jul10-12, 04:15 PM   #32

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Quote by roovid View Post
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??? Even the problem maker can not assume nonsense....

ehild
Jul10-12, 04:17 PM   #33
 
lol i kno
but some of the question i am doin are makin me assume some weird stuff
for basic calculation purposes
sorry.my error
thnx fr the correction again
Jul10-12, 04:19 PM   #34

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You are welcome.

ehild
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