shell52080
I'm having some trouble with this physics problem.

Here's the question:

A rocket, initially at rest is fired vertically with an upward acceleration of 10m/s2. At an altitude of .5km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

First I found the time to be 10s. Then I found the velocity from the time to be 100m/s. From there I solved for y and got 510m. Is this correct?
amac
Close. You need to add the altitude the rocket already had when the engines cut out though, so it would be 510m + 500m = 1.01 km.
shell52080
Thank you!
Grasshopper
Can someone explain in detail how you got your answers? I've never had to solve a problem where an engine cuts off in mid- assent (I've only done problems involving, say, someone throws up a rock, etc.)... but

Is it possible to use anti-derivatives to solve this problem?

If acceleration is given, then the anti-derivative of acceleration is velocity, and the anti-derivative of velocity is displacement...

So, when the maximum height is reached, then velocity will be zero, right? So, couldn't you set velocity equal to zero and then solve for t, and then plug t into the displacement function to find out the height when velocity is equal to zero (when the rocket reaches its maximum height)...

However, I am aware that I am not including the fact that the engines shut off at .5 km... so...

Could someone please explain in detail how to solve that problem?

Thanks
mr_homm
You can of course use antiderivatives (integrals) to solve kinematic problems in general, but for the present case, that is unnecessary.

All you must remember is that the equations of kinematics are valid when the acceleration is CONSTANT, but any change in acceleration means that you cannot use them. Therefore, if you can break the motion up into STAGES, each with a constant acceleration, you can solve each stage using kinematics. The ending position and velocity of each stage are the beginning position and velocity for the next stage.

In this way, a multistage kinematics problem is really no harder than several ordinary kinematics problems strung together. You just do them one at a time, from start to finish, using the acceleration that is valid at each stage.

In the rocket problem, you have the initial velocity at the ground, and the acceleration while the engine fires, and the height at which the engine shuts off. Use kinematics to find the velocity of the rocket at the moment the engine shuts off. Then the rocket height and velocity are the starting point for stage 2. In stage 2, the acceleration is just gravity, -9.81m/s^2, because the engine is off. The starting height and velocity are known, and the ending velocity is zero at the maximum height. Use kinematics again to find the final height.

Hope this helps!

--Stuart Anderson
Grasshopper
QUOTE (mr_homm+Mar 1 2007, 09:36 PM)
You can of course use antiderivatives (integrals) to solve kinematic problems in general, but for the present case, that is unnecessary.

All you must remember is that the equations of kinematics are valid when the acceleration is CONSTANT, but any change in acceleration means that you cannot use them.  Therefore, if you can break the motion up into STAGES, each with a constant acceleration, you can solve each stage using kinematics.  The ending position and velocity of each stage are the beginning position and velocity for the next stage.

In this way, a multistage kinematics problem is really no harder than several ordinary kinematics problems strung together.  You just do them one at a time, from start to finish, using the acceleration that is valid at each stage.

In the rocket problem, you have the initial velocity at the ground, and the acceleration while the engine fires, and the height at which the engine shuts off.  Use kinematics to find the velocity of the rocket at the moment the engine shuts off.  Then the rocket height and velocity are the starting point for stage 2.  In stage 2, the acceleration is just gravity, -9.81m/s^2, because the engine is off.  The starting height and velocity are known, and the ending velocity is zero at the maximum height.  Use kinematics again to find the final height.

Hope this helps!

--Stuart Anderson

Well, that IS informative, and thank you. But the problem is, I have not taken a physics course yet (next fall I will take calculus based physics, but they make us take two chemistry courses first at my school).

Would it be too much trouble to ask for an explanation of these equations of kinematics? Or could you direct me elsewhere in the forums?

Thanks
mr_homm
Hi Grasshopper,

Sorry, I didn't know you hadn't had physics yet. I gave the kinematics equations to someone else just a few days ago, so here's a link to that post. (By the way, that other person posted the question in physics/general, but that wasn't really the right place for homework questions. The forum you're in now is the right one.)

If you combine the equations and advice I gave there with the method I gave in this thread, you should be able to solve this problem easily. Let me know if you have any further questions.

Hope this helps!

--Stuart Anderson
Grasshopper
QUOTE (mr_homm+Mar 3 2007, 02:32 PM)
Hi Grasshopper,

Sorry, I didn't know you hadn't had physics yet. I gave the kinematics equations to someone else just a few days ago, so here's a link to that post. (By the way, that other person posted the question in physics/general, but that wasn't really the right place for homework questions. The forum you're in now is the right one.)

If you combine the equations and advice I gave there with the method I gave in this thread, you should be able to solve this problem easily. Let me know if you have any further questions.

Hope this helps!

--Stuart Anderson

Thanks alot! That is quite enlightening.
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