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LT72884

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- Finding max height for a rocket, then terminal velocity during free fall

Ok, so here is what i am trying to do. Find apogee of a model rocket i built using a C motor, then i need to calculate terminal velocity, but something is not adding up, i get a really really high apogee, almost a mile.

STAGE 1:

I have a c6-7 rocket motor which provides 10N-s of impulse, and 6N of thrust.

Burn time = Impulse/Thrust = 10/6 = 1.66 seconds of burn time

Rocket mass is 53g or 0.05398kg

Weight = 0.529N

a=F/m where F=Th-mg so Resultant force is 5.47N

acceleration =101.35m/ss

Burnout Velocity = v0+at = 0+(101.35)(1.66)=168.2m/s

Height of rocket after burnout velocity is y1=v0t+(at^2)(0.5) = 0+(101.35)(1.66^2)/2=139m (seems reasonable for a C motor)

STAGE 2: after burnout Acceleration is now -9.8

vi=168.241m/s

at top of apogee, v=0 so 0=v0+at t=-v0/a t=(-168.241)/-9.8 = 17.1 seconds of coasting time. this seems like ALOT OF TIME maybe because rocket is so light??

now y2 = v0t-at^2(0.5) = 168.241)(17.167)-((9.8)(17.167^2)(0.5)= 1444m

total apogee is y1+y2 = 1583m or 4749feet give or take...This seems ridiculous and un-realistic, maybe because a C motor is way to powerful for that size of rocket?

Ok, let's say these numbers do indeed work out, how do i find terminal velocity now without drag? is it sqrt(2gh)? I am trying to find Cd myself to compare with my win tunnel results by using

Cd=((2mg)/(rho(vt^2)(Area)

EDIT: i just looked in my fluid dynamics book, and i have a feeling that to find the CD or drag force without testing is a bunch of crazy integration?

thanks

STAGE 1:

I have a c6-7 rocket motor which provides 10N-s of impulse, and 6N of thrust.

Burn time = Impulse/Thrust = 10/6 = 1.66 seconds of burn time

Rocket mass is 53g or 0.05398kg

Weight = 0.529N

a=F/m where F=Th-mg so Resultant force is 5.47N

acceleration =101.35m/ss

Burnout Velocity = v0+at = 0+(101.35)(1.66)=168.2m/s

Height of rocket after burnout velocity is y1=v0t+(at^2)(0.5) = 0+(101.35)(1.66^2)/2=139m (seems reasonable for a C motor)

STAGE 2: after burnout Acceleration is now -9.8

vi=168.241m/s

at top of apogee, v=0 so 0=v0+at t=-v0/a t=(-168.241)/-9.8 = 17.1 seconds of coasting time. this seems like ALOT OF TIME maybe because rocket is so light??

now y2 = v0t-at^2(0.5) = 168.241)(17.167)-((9.8)(17.167^2)(0.5)= 1444m

total apogee is y1+y2 = 1583m or 4749feet give or take...This seems ridiculous and un-realistic, maybe because a C motor is way to powerful for that size of rocket?

Ok, let's say these numbers do indeed work out, how do i find terminal velocity now without drag? is it sqrt(2gh)? I am trying to find Cd myself to compare with my win tunnel results by using

Cd=((2mg)/(rho(vt^2)(Area)

EDIT: i just looked in my fluid dynamics book, and i have a feeling that to find the CD or drag force without testing is a bunch of crazy integration?

thanks

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