2d Kinematics: rocket fired up, engine dies

In summary: Yf = 0 + 0 + 1/2(5)(23.52^2)Yf = 1382.98mis that right?No, I used wrongly the value 23.52 seconds you calculated. The value of a is 5 m/s^2 from t = 0 until an unknown time t = t_1, when the velocity will reach a maximum value V_fWe know that after some time t_f = t_1 + t_2 the velocity will be zero.Then, the rocket will fall, accelerated by gravity, and after a time t_3, will hit the ground.We know that t_2 + t_3 =
  • #1
blumatrix
3
0
?: 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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  • #2
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
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
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
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
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
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
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]
 

1. What is 2D kinematics?

2D kinematics is the study of the motion of objects in two-dimensional space, taking into account both the horizontal and vertical components of motion.

2. How does a rocket's engine dying affect its motion in 2D kinematics?

If a rocket's engine dies while it is in motion, it will continue to move in a straight line with a constant velocity, but its speed and direction may change depending on external forces such as gravity and air resistance.

3. How can we calculate the rocket's position and velocity at any given time in 2D kinematics?

We can use equations of motion, such as the kinematic equations, to calculate the rocket's position and velocity at any given time. These equations take into account factors such as initial velocity, acceleration, and time.

4. What factors affect the motion of a rocket in 2D kinematics?

The motion of a rocket in 2D kinematics is affected by factors such as initial velocity, acceleration due to gravity, air resistance, and any external forces acting on the rocket.

5. How can we analyze the motion of a rocket in 2D kinematics graphically?

We can plot the rocket's position, velocity, and acceleration as a function of time on a graph to analyze its motion in 2D kinematics. This allows us to see any changes or patterns in its motion over time.

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