Rocket Acceleration: Calculating Speed and Altitude

AI Thread Summary
A 1000kg weather rocket accelerates straight up for 16 seconds before the motor stops, reaching an altitude of 5100m after 20 seconds with no air resistance. To find the rocket's acceleration during the first 16 seconds, the kinematic equation r = Vit + 0.5a(t^2) is suggested. After the motor cuts off, the rocket experiences gravitational acceleration of -9.8 m/s² for the last 4 seconds, leading to a maximum height. The velocity at engine cutoff can be calculated using the known final velocity of 0 m/s at the peak height. This approach allows for determining both the rocket's acceleration and speed as it passes through the cloud at 5100m.
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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