Rocket Acceleration: Calculating Speed and Altitude

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SUMMARY

The discussion focuses on calculating the acceleration and speed of a 1000kg weather rocket launched vertically. The rocket experiences a constant acceleration for the first 16 seconds, after which the motor stops. The altitude reached after 20 seconds is 5100m, allowing for the application of kinematic equations to determine the rocket's acceleration and speed at that altitude. The key equations used include r = Vi*t + 0.5*a*t² and the understanding that gravitational acceleration (g = -9.8 m/s²) affects the rocket after the motor cutoff.

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mullets1200
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Homework Statement



A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. There is no air resistance.

Questions:
What was the rocket's acceleration during the first 16s?

What is the rocket's speed as it passes through a cloud 5100m above the ground
 
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mullets1200 said:

Homework Statement



A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. There is no air resistance.

Questions:
What was the rocket's acceleration during the first 16s?

What is the rocket's speed as it passes through a cloud 5100m above the ground

Equations?

And your attempts with them?
 
pgardn said:
Equations?

And your attempts with them?


Equations:
I think you use the equation: r=Vit+.5a(t)2

Im not sure where to start though that is the problem
 
mullets1200 said:
Equations:
I think you use the equation: r=Vit+.5a(t)2

Im not sure where to start though that is the problem

You are going to need your other kinematic equations.

I would start by looking at the last 4 seconds of the rockets trip (when the only force acting on the rocket is gravity thus a = g = -9.8m/s/s) because the engine has cut off. So when the engine cuts off at 16 seconds you have reached your maximum velocity. After that, during the last 4 seconds the rocket will be slowing down (accelerating down). At the end of this 4 seconds the rocket will be going 0 m/s presumably as it has reached its maximum height.

So why not find the velocity of the rocket when the engine cut off. You got a = g, you got t, and you got Vf = 0 m/s ... find Vo which will be the velocity when the engine cuts off...

Then look at the first 16 s during which the rocket is accelerating up. You have t, you have Vf from the above (it is really Vo from the above), you have Vo = 0 m/s (I will assume the rocket started from rest from the lauch pad) and solve for a.

This is one way to go about it.
 

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