Rocket firing problem

1. Jan 23, 2004

dorthod

Did I solve this correctly?

A 0.4-kg model rocket is launched straight upwards. Its engine provides an impulse of 22 N*s by firing for 3 seconds. What is the average acceleration of the rocket while the engine is firing, and what is the maximum height the rocket will reach?

J=Ft so 22=F(3) so F=7.33
F=ma so 7.33=4a so a=1.83
x=(1/2)at(squared) so x=(1/2)1.83(3)squared so x=8.25 meters
then the engines cut off
the final velocity = at so v=(1.83)(3) so v=5.5 m/s
then using v(squared)=v0(squared) + 2a(x-x0) to find the displacement at the top of the arch I get 0 = 5.5(squared) + 2(-9.8)(x-8.25) so x = 9.79m

My two answers would be 1.83 m/s(squared) and 9.79 meters, correct or not?

2. Jan 23, 2004

Jimmy

Check your value for mass. It was given as .4Kg. You used 4Kg in your calculation.

Also, the acceleration of gravity is working in the opposite direction as the acceleration provided by the rocket motor.

other than that, your method seems fine.

Last edited: Jan 23, 2004
3. Jan 23, 2004

dorthod

doh, I'm an idiot. Using the same method as below, with .4 as the mass: So 7.33/.4 = 18.333 which is a, but minus gravity, the acceleration of the rocket is 8.53. That means it was at 38.4m and going 25.6 m/s when the engines cut off. Therefore it was 71.837 m high at the peak. Correct?

4. Jan 23, 2004

Jimmy

Don't be to hard on yourself. It's easy to forget things like that.