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Rocket Launch in 3 stages

  1. Feb 15, 2009 #1
    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
    [tex]\Delta[/tex]t=0.01s

    Unknowns:
    acceleration
    velocity
    height
    total time

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

    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
    Fe-w=may
    ay=[tex]\frac{Fe-w}{m}[/tex]
    ay=[tex]\frac{16N-(0.05)(9.81)}{0.05}[/tex]=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 [tex]\leq[/tex] t [tex]\leq[/tex] 6.15s
    [tex]\Sigma[/tex]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 [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.
     
  2. jcsd
  3. Feb 15, 2009 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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?
     
  4. Feb 15, 2009 #3
    So I would be using -9.81 instead of 9.81 for gravity now? Also would apex be the addition of the height of stage 1 & 2 added together. 111.11 + 10.47 = 121.58m?

    Also the velocity and height numbers seem small given the acceleration at lift off.
     
    Last edited: Feb 15, 2009
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