2d Kinematics: rocket fired up, engine dies

  • Thread starter blumatrix
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  • #1
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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.
 

Answers and Replies

  • #2
radou
Homework Helper
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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.
 
  • #3
SGT
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]
 
  • #4
radou
Homework Helper
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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.
 
  • #5
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initial velocity can be 0 if the object is not in motion, right?

and 5 m/s^2 is the acceleration of the rocket.
 
  • #6
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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?
 
  • #7
SGT
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].
 
  • #8
SGT
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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