2d Kinematics: rocket fired up, engine dies

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Homework Help Overview

The discussion revolves around a kinematics problem involving a rocket that is fired upward with an initial acceleration of 5 m/s². After a certain time, the engine fails, and the rocket enters freefall, hitting the ground 12 seconds later. Participants are tasked with finding the final velocity and maximum height of the rocket.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the rocket's initial velocity, acceleration, and the time of flight. There are attempts to calculate the height and final velocity using kinematic equations, with some questioning the initial conditions and assumptions about the motion of the rocket after the engine fails.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided equations and calculations, while others have raised questions about the initial conditions and the application of kinematic equations. There is no explicit consensus on the correct approach or final values yet.

Contextual Notes

Participants are navigating the complexities of the problem, including the transition from powered flight to freefall and the implications of the rocket's motion during these phases. There are mentions of specific time intervals and the need to clarify the initial conditions for accurate calculations.

blumatrix
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?: A rocket is fired up at 5 m/s^2. After a certain time, the engine fails and the rocket goes into freefall. It hits the ground 12 seconds after engine failing. Find V(final) and height that the rocket reaches.

the the way down...
Vf = Vi - at
set Vf = 0 b/c when it hits the ground, V = 0
Vi = 9.8(12)
Vi = 117.6 m/s

so plugging this back into the trip up...
Vf = Vi - at
Vf = 117.6 m/s
Vi = 0 b/c it starts at 0 before fired.
117.6 m/s = - (5 m/s^2)t
t = 23.52 sec

i hope this is right so far... so how do i find V(final) and the height using this? the rocket doesn't stop when the engine dies. it continues up for a little bit before starting the freefall...

thx guys.
 
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blumatrix said:
?: A rocket is fired up at 5 m/s^2. After a certain time, the engine fails and the rocket goes into freefall. It hits the ground 12 seconds after engine failing. Find V(final) and height that the rocket reaches.

the the way down...
Vf = Vi - at
set Vf = 0 b/c when it hits the ground, V = 0
Vi = 9.8(12)
Vi = 117.6 m/s

so plugging this back into the trip up...
Vf = Vi - at
Vf = 117.6 m/s
Vi = 0 b/c it starts at 0 before fired.
117.6 m/s = - (5 m/s^2)t
t = 23.52 sec

i hope this is right so far... so how do i find V(final) and the height using this? the rocket doesn't stop when the engine dies. it continues up for a little bit before starting the freefall...

thx guys.

Did you perhaps mean 'A rocket is fired up at 5 m/s' ?

Vi = 0 b/c it starts at 0 before fired. -> What exactly did you mean by that? The initial velocity can not equal zero.
 
blumatrix said:
?: A rocket is fired up at 5 m/s^2. After a certain time, the engine fails and the rocket goes into freefall. It hits the ground 12 seconds after engine failing. Find V(final) and height that the rocket reaches.

the the way down...
Vf = Vi - at
set Vf = 0 b/c when it hits the ground, V = 0
Vi = 9.8(12)
Vi = 117.6 m/s

so plugging this back into the trip up...
Vf = Vi - at
Vf = 117.6 m/s
Vi = 0 b/c it starts at 0 before fired.
117.6 m/s = - (5 m/s^2)t
t = 23.52 sec

i hope this is right so far... so how do i find V(final) and the height using this? the rocket doesn't stop when the engine dies. it continues up for a little bit before starting the freefall...

thx guys.
The heigth at any instant t can be calculated by:
[tex]h = h_0 + v_0\cdott + \frac{1}{2}\cdota\cdott^2[/tex]
The maximum height is reached when V = 0.
From 0 to 23.52 seconds, [tex]a = 5 ms^{-2}[/tex]
After that time [tex]a = -g = -9.8 ms^{-2}[/tex]
 
SGT said:
The heigth at any instant t can be calculated by:
[tex]h = h_0 + v_0\cdott + \frac{1}{2}\cdota\cdott^2[/tex]
The maximum height is reached when V = 0.
From 0 to 23.52 seconds, [tex]a = 5 ms^{-2}[/tex]
After that time [tex]a = -g = -9.8 ms^{-2}[/tex]

That would be the height at t = 1, not at any instant t.
 
initial velocity can be 0 if the object is not in motion, right?

and 5 m/s^2 is the acceleration of the rocket.
 
so velocity will be 0 at the peak after the engine dies...
setting V = 0, and using Y(final) = Yi + Vi(t) + 1/2at^2
Yf = 0 + 0 + 1/2(5)(23.52^2)
Yf = 1382.98m

is that right?
 
blumatrix said:
so velocity will be 0 at the peak after the engine dies...
setting V = 0, and using Y(final) = Yi + Vi(t) + 1/2at^2
Yf = 0 + 0 + 1/2(5)(23.52^2)
Yf = 1382.98m

is that right?
No, I used wrongly the value 23.52 seconds you calculated.
The value of a is [tex]5 m/s^2[/tex] from t = 0 until an unknown time [tex]t = t_1[/tex], when the velocity will reach a maximum value [tex]V_f[/tex]
We know that after some time [tex]t_f = t_1 + t_2[/tex] the velocity will be zero.
Then, the rocket will fall, accelerated by gravity, and after a time [tex]t_3[/tex], will hit the ground.
We know that [tex]t_2 + t_3 = 12 s[/tex]
You have 4 unknowns: [tex]t_1, t_2, t_3, V_f[/tex] and 3 equations of movement: from 0 to [tex]t_1[/tex], from [tex]t_1[/tex] to [tex]t_f = t_1 + t_2[/tex] and from that time to [tex]t_f + t_3[/tex] and an equation of time: [tex]t_2 + t_3 = 12 s[/tex].
 
radou said:
That would be the height at t = 1, not at any instant t.
I don't know what happened with my LaTex equation. It should read:
[tex]h = h_0 + V_0t + \frac{1}{2}at^2[/tex]
 

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