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1D Kinematics question

  1. Dec 19, 2015 #1
    1. The problem statement, all variables and given/known data

    During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth’'s surface and is to reach a maximum height of 990 m above the earth'’s surface. The rocket’'s engines give the rocket an upward acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance.

    What it asks for : What must be the value of T in order for the rocket to reach the required altitude?



    2. Relevant equations
    basic kinematics formulas
    X= Xo + VoT+1/2AT^2
    V=Vo+AT
    V^2=Vo^2+2A(X-Xo)

    o denotes initial


    3. The attempt at a solution
    | X3= 990 V3=0
    |
    |
    | X2= V2= A=-9.8
    |
    |
    |
    | X1=0 V1=0 A=16.0 m/s

    X3=X2+V2T-1/2 9.8T^2
    X2=X1+V1T+.5*16*T^2
    X2=0+0T+.5^16*T^2
    X2=8T^2

    plugging pack into original equation:
    X3=8T^2 +V2T -4,9 T^2
    combining like terms
    X3= 3.1T^2 + V2T

    finding V2:
    V2=V1+16T
    V2=0+16T
    V2=16T
    plugging back in

    X3= 3.1T^2 + 16T
    X3=990
    990=3.1T^2 + 16T

    it is a quadratic equation I graphed it to find the X value, I got 20.65 , but it is wrong, where Did I go wrong?
     
  2. jcsd
  3. Dec 19, 2015 #2
    The 'T's that appear here
    and
    are not the same.
    The first is the time it takes for the rocket to reach zero velocity after the engines were shut off, while the second is the boost time during which the engines were turned on.
     
  4. Dec 19, 2015 #3
    So what alternative method should I use?
     
  5. Dec 19, 2015 #4
    There is no need for an alternative method; your general approach is okay - just that you need to be careful when you keep track of the two different times. To proceed, you have to find out how the time [itex]t_{2}[/itex] in [itex]x_{3}=x_{2}+ v_{2}t_{2} -\frac{1}{2} g {t_{2}}^2[/itex] is related to the boost time [itex]T[/itex].
    (Hint: Use [itex]v=v_{0}+at[/itex])
     
  6. Dec 19, 2015 #5
    V3=V2+AT
    V2=16t
    V3=16t-9.8T
    V3=0
    0=16t-9.8T
    -16t=-9.8T
    16/9.8t = T

    8t2+16t2-4.9T2
    8t2+16t2-4.9(16/9.8 t)2

    990= 10.93 t2
    90.5=t2
    √90.5 =t
    t= 9.513

    the correct answer is 6.86.
     
  7. Dec 19, 2015 #6
    Yup, that's what I got.

    This line is not correct:
    It should be
    [tex]x_{3} = x_{2} + v_{2}t - \frac{1}{2}g t^2 = \frac{1}{2}a T^{2} + (aT)t - \frac{1}{2}g t^2[/tex]
    Substitute [itex]t = (a/g)T [/itex] as you've derived above, and you should be done.

    (Note: I realised the meaning of t and T I've used here is opposite of what you have in that post, but you should be able to make the connection :p)
     
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