How Does Gravity Affect a Model Rocket's Upward Acceleration?

AI Thread Summary
The discussion centers on understanding the effects of gravity on a model rocket's upward acceleration and comparing it to other scenarios. A shuttle bus decelerates to a stop, resulting in a final velocity of 0 m/s, while a camera dropped from a cliff has a non-zero final velocity just before impact, as it is still moving when it hits the ground. The model rocket's upward acceleration of 2.00 m/s² must be considered alongside gravitational acceleration of 9.81 m/s², leading to a net upward acceleration. Participants clarify that the final velocity of the camera refers to its speed just before impact, contrasting with the gradual stop of the bus. The conversation emphasizes the importance of understanding these dynamics in physics.
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1. A shuttle bus slows down with an average acceleration of -1.8m/s^2. How long does it take the bus to slow from 9.0m/s to a complete stop? For this question, we can immediately say that the final velocity is 0m/s because it stops. Now let's take a look at the second question: A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the velocity with which the camera hits the ground. Why here this final velocity is not '0' m/s because i think it hits the ground, so it stops?
2. A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150m. My question is its acceleration is 2.00m/s^2 upward, but do we need to consider about the gravity 9.81m/S^2 that pulls it down because that's the reason why later on it will slow down?
Thanks a lot for explanations.
 
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For what I presume is the first question, the final velocity is the maximum velocity reached before being accelerated in the opposite direction by the normal force. For the second question, I think you can assume that the acceleration due to gravity was already taken into account by the 2.oo ms^{-2}, giving a total upward acceleration of 11.81 ms^{-2}, assuming the acceleration due to gravity does not change to a significant degree over the 150m.
 
MIA6 said:
1. A shuttle bus slows down with an average acceleration of -1.8m/s^2. How long does it take the bus to slow from 9.0m/s to a complete stop? For this question, we can immediately say that the final velocity is 0m/s because it stops. Now let's take a look at the second question: A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the velocity with which the camera hits the ground. Why here this final velocity is not '0' m/s because i think it hits the ground, so it stops?
2. A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150m. My question is its acceleration is 2.00m/s^2 upward, but do we need to consider about the gravity 9.81m/S^2 that pulls it down because that's the reason why later on it will slow down?
Thanks a lot for explanations.

homework questions in homework forums
 
It's not my homework question, it's just my classwork that i still didn't really understand, so i posted it and asked other people.
 
MIA6 said:
It's not my homework question, it's just my classwork

Semantics will get you nowhere, my friend.
 
MIA6 said:
It's not my homework question, it's just my classwork that i still didn't really understand, so i posted it and asked other people.

Please re-read the Forum Guidelines section on Homework Help. Coursework is included, and needs to be posted in the Homework Help forums, not in the general technical forums.

https://www.physicsforums.com/showthread.php?t=5374

Now, using bel's help so far, can you please show us some of your work on the questions, and we'll try to provide some tutorial help when you get stuck.
 
Ok,i will post it in homework help next time. I am a new member, so i don't really know the rules. berkeman, thanks for reminding me. These two questions i have already solved, but i don't understand some details. For question 1, I still can't figure out in second example why the final velocity is not 0 as in the first one.
 
MIA6 said:
For question 1, I still can't figure out in second example why the final velocity is not 0 as in the first one.

Do you mean the rocket one? It has a Vo, and its final velocity is that plus the extra velocity it gains due to the 2m/s^2 net acceleration. The 2m/s^2 is the actual acceleration, which is what it could do without gravity, minus the 9.8m/s^2 gravitational acceleration that it is having to fight. Does that help?
 
It helps, but i mean the first question about comparing the final velocity of the shuttle bus and of camera dropped from Mars. For the bus one, we know that the Vf is 0 because it comes to stop, but why the Vf of camera hits the ground is not 0? It hits the ground means it stops?
 
  • #10
Ah, gotcha. When they say "the final velocity with which it hits the ground", they mean just barely before impact. The force of the impact will bring the velocity down to zero a short time later, but that's not what they are asking about. The train slowly decelerated to zero (smoothly), so that's why it's final velocity is zero (or infinitesimally small just before it "stops"), as opposed to the camera which has full velocity just before it "stops".
 
  • #11
So it means the final velocity before it hits the ground, right? I thought of that, too. Btw, when the camera hits the ground, it's not gradual like the train because train slows gradually until it stops, right? but camera is like suddenly stops?
 
  • #12
MIA6 said:
So it means the final velocity before it hits the ground, right? I thought of that, too. Btw, when the camera hits the ground, it's not gradual like the train because train slows gradually until it stops, right? but camera is like suddenly stops?

Correctomundo. Sounds like you have it. :smile:
 
  • #13
berkeman, thanks a lot for help.
 
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