# Rocket Launch in 3 stages

1. Feb 15, 2009

A brief explanation, this assignment is to be done in Excel and also has to be turned in with handwritten solutions for finding total flight time and apex. If I can get the handwritten part done Excel is a piece of cake.
1. The problem statement, all variables and given/known data
A model rocket is launched and I am to evaluate the 3 stages it goes through. During the first 0.15s the rocket (m=0.05 kg) is launched with a force of 16N. It then coasts upward while being slowed done by gravity (g=9.81m/s2. After it reaches apex(max altitude it starts to fall back down. It also deploys a parachute 6 sec. after the motor stops with a constant speed of 10 m/s until it hits ground. Calculate the speed and altitude of the rocket over it's flight time and plot them.

Knowns:
g=9.81m/s2
FE=16N
m=0.05 kg
vo=0m/s
$$\Delta$$t=0.01s

Unknowns:
acceleration
velocity
height
total time

2. Relevant equations
$$\Sigma$$F=ma
v(t)=vo+at
h(t)=ho+vo+.5at2

3. The attempt at a solution

Stage 1: 0$$\leq$$ t $$\leq$$ 0.15s
use dt=0.01s (was told to do this by teacher for stage 1)
$$\Sigma$$F=ma
Fe-w=may
ay=$$\frac{Fe-w}{m}$$
ay=$$\frac{16N-(0.05)(9.81)}{0.05}$$=310.19m/s2

v=vo+at
v=0+(310.19)(0.15)=46.53m/s

h=ho+v0+.5at2
h=.5at2
h=.5(310.19)(0.15)2
h=10.47 m

Stage 2: 0.15 $$\leq$$ t $$\leq$$ 6.15s
$$\Sigma$$F=ma
-w=may
ay=-g

v=vo+at
v=46.53+(9.81)(6.15)=106.86m/s

h=10.47m+(46.53)(6.15)+.5(-9.81)(6.15)2=111.11m

Stage 3: 6.15 $$\leq$$ t $$\leq$$ total time

Stage 3 is were I get lost could some point me in the right direction as to find total time and apex.

2. Feb 15, 2009

### D H

Staff Emeritus
Your stage 2 is incorrect. At the end of stage 1, you have the rocket's velocity as 46.53 m/s upwards (correct). Then at the end of stage 2, during which the only force is gravity (downward), you have the rocket's velocity as 106.86 m/s upwards. Ask yourself, does this make any sense?

3. Feb 15, 2009