Rocket's Max Altitude: 295 kg Weather Rocket

[sktgurl930]

A 295 kg weather rocket is loaded with 112 kg of fuel and fired straight up. It accelerates upward at 33.67 m/s2 for 35.2 s, then runs out of fuel. Ignore any air resistance effects.
What is the rocket's maximum altitude?

Im so lost if someone can please help me it would be so awesome

 

[LowlyPion]

A 295 kg weather rocket is loaded with 112 kg of fuel and fired straight up. It accelerates upward at 33.67 m/s2 for 35.2 s, then runs out of fuel. Ignore any air resistance effects.
What is the rocket's maximum altitude?

Im so lost if someone can please help me it would be so awesome

The rocket requires 2 calculations.

The first is to determine the speed and height it gets to during the rocket ignition phase.

Thereafter it is being slowed by gravity. Determine from the speed you calculated as a result of the rocket ignition how much higher it will go. Add the 2 distances together.

Here are some formulas to help you.
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[RoryP]

hmmm not too sure. have you considered any SUVAT equations?? S=ut + 1/2at^2 perhaps then something to do with G.P.E??

 

[sktgurl930]

ok i think I am doing something wrong here cause i got V= 1185.184, and X to = 41718.4768

so with the displacement formula i put x= V*35.2+.5*33.67*T^2

and my answer was 62577.7152 m

 

[LowlyPion]

ok i think I am doing something wrong here cause i got V= 1185.184, and X to = 41718.4768

so with the displacement formula i put x= V*35.2+.5*33.67*T^2

and my answer was 62577.7152 m

Your first answer is incorrect. The height will be given simply by 1/2 a* t ² and that looks like you are missing a factor of 1/2. Your initial velocity is correct.

On the second part you have x from the Vf² - vi² = 2 a*x. For Vi = 1185.184
(At max height vf will be = 0.)

x = (1185.184)²/2(9.8)

Then add the corrected value of the first part and you're done.

 

[DaveC426913]

The amount of fuel is listed as a given. One must factor in changing mass and therefore acceleration.

 

[LowlyPion]

The amount of fuel is listed as a given. One must factor in changing mass and therefore acceleration.

Given the statement of the problem, with constant acceleration given, the rate of change of mass should not be a concern. The fuel is apparently expended as it enters the gravity only phase is all you apparently need to know.

If the statement of the problem indicated that initial acceleration was xxx and it maintained constant force, then I agree it would be a concern. But I don't read it that way. And since I read this as an introductory question, I think the simpler interpretation is the most likely.

 

[DaveC426913]

Given the statement of the problem, with constant acceleration given, the rate of change of mass should not be a concern. The fuel is apparently expended as it enters the gravity only phase is all you apparently need to know.

If the statement of the problem indicated that initial acceleration was xxx and it maintained constant force, then I agree it would be a concern. But I don't read it that way. And since I read this as an introductory question, I think the simpler interpretation is the most likely.

You're right. Upon re-reading, I see that the acceleration is also a given, and is constant.