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.(adsbygoogle = window.adsbygoogle || []).push({});

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/s^{2}. 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/s^{2}

F_{E}=16N

m=0.05 kg

v_{o}=0m/s

[tex]\Delta[/tex]t=0.01s

Unknowns:

acceleration

velocity

height

total time

2. Relevant equations

[tex]\Sigma[/tex]F=ma

v(t)=v_{o}+at

h(t)=h_{o}+v_{o}+.5at^{2}

3. The attempt at a solution

Stage 1: 0[tex]\leq[/tex] t [tex]\leq[/tex] 0.15s

use dt=0.01s (was told to do this by teacher for stage 1)

[tex]\Sigma[/tex]F=ma

F_{e}-w=may

a_{y}=[tex]\frac{Fe-w}{m}[/tex]

a_{y}=[tex]\frac{16N-(0.05)(9.81)}{0.05}[/tex]=310.19m/s^{2}

v=v_{o}+at

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

h=h_{o}+v_{0}+.5at2

h=.5at^{2}

h=.5(310.19)(0.15)2

h=10.47 m

Stage 2: 0.15 [tex]\leq[/tex] t [tex]\leq[/tex] 6.15s

[tex]\Sigma[/tex]F=ma

-w=ma_{y}

a_{y}=-g

v=v_{o}+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 [tex]\leq[/tex] t [tex]\leq[/tex] total time

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

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# Rocket Launch in 3 stages

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