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Launch a weather rocket prbm

  1. Jun 20, 2007 #1

    ere

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    "launch a weather rocket" prbm

    1. The problem statement, all variables and given/known data
    A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 11 s, then the motor stops. The rocket altitude 15 s after launch is 4600 m. You can ignore any effects of air resistance.

    a) Acc. of rocket during the first 11s
    b) Its speed at 15s

    2. Equations
    ------------
    V(final)^2 = V(init.)^2 + 2AS
    -> therefore S = V(final)^2 / (2A) when V(init.) = 0
    ------------
    S(final) = S(init.) = V(init.)t + .5At^2
    ------------

    3. The attempt at a solution
    V(@11s) = S(init.) + V(init.)t + .5At^2
    = 0 + 0t + .5a(11)^2 = 60.5a
    thus S(first 11s) = V(@11s)^2 / 2a as V(init.) = 0
    = (60.5a)^2 / 2a = 1830.125a
    S(total) = 4600 = S(first 11s) + V(@11s)(4) + .5(0)t^2
    4600 = 1830.125a + 60.5a(15-11) + 0 = (1830.125 + 242)a = 2072.125a
    therefore a = 2.21994

    and the homework system returns wrong. i hope someone can point out where i went wrong. thanks!
     
  2. jcsd
  3. Jun 20, 2007 #2

    G01

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    You are making part A a tiny bit harder than it has to be. You know the acceleration caused by the rocket. Does gravity accelerate the rocket? If so, what would be the total acceleration?

    If you get part A, then part B is a kinematics problem. HINT: Remember that the total acceleration changes at 11s. You may need to split up the problem into two parts.

    See how far you can get now. If you need more help, feel free to ask. Good Luck!
     
  4. Jun 20, 2007 #3

    ere

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    the gravity put an additional -9.8ms-2 to the acceration, so should i assume:
    1. the acceration of the rocket's motor is actually a+9.8 in order to overcome the effect of the gravity? then the answer is 2.220+9.8 = 12.02 which is wrong according to the system.
    2. when the motor stops, the only force acting on the rocket is the grativational acceration. so the rocket is actually descending?
     
  5. Jun 20, 2007 #4

    G01

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    Your second assumption is correct. I think I misread the question. I thought you were given the rocket engine's acceleration. Sorry about this!!

    OK. Here's what you have.

    You know the rocket's position at 15s. and You know it's acceleration between 11s and 15s, since only gravity is acting on the rocket.

    Call its position at 11s "y". Can you set up two equations involving y? The first one should describe the situation from 0-11s. The second should describe the situation from 11-15s. If you can do this, would you then be able to find a, the acceleration between 0 and 11s?
     
  6. Jun 20, 2007 #5

    ere

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    V(@11s)^2 = V(init.)^2 + 2AS(11s)
    S(11s) = sqrt((V(@11s)^2 - V(init.)^2) / 2A)
    = sqrt((V(@11s)^2 - 0) / 2(a - 9.8))
    ^ that will be the distance travelled during the first 11 seconds, with gravitational effect, in terms of total acceleration

    S(15s) = S(11s) + V(@11s)t + .5At^2
    = (the whole thing from equation 1) + V(@11s, in terms of a)(15-11) + .5(-9.8)(4^2)
    this way, the last part of the 2nd equation concerning distance travelled during 11 to 15 seconds is negative.
    sorry if i am making this way to complicated for the reader.
    or, for part 1: S(11s) = S(init.) + V(init.)11 + 2(a-9.8)(11^2)
    = 0 + 0 + 242a - 2371.6 = 242a - 2371.6
     
  7. Jun 20, 2007 #6

    ere

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    ok that was quite a mess.
    V(@11s) = V(init.) + At = 0 + (a - 9.8)11 = 11a - 107.8

    therefore the distance the rocket fell after motor has stopped (ie ignoring distance already travelled) is:
    S = S(11s, ignore) + V(@11s)t + .5(-9.8)t^2
    = 0 + (11a - 107.8)(4) + (-9.8)(8)
    = 44a - 431.2 - 78.4 = 44a - 509.6

    thus the distance travelled before motor stopped is 4600 + 44a - 509.6

    so, the distance travelled with the motor on (first 11sec) is
    S = S(init) + V(init)t + .5At^2
    = 0 + 0t + .5(a - 9.8)11^2 = 4600 + 44a - 509.6
    = 60.5(a - 9.8) = 4090.4 + 44a
    = 60.5a - 592.9 = 44a + 4090.4
    = 16.5a = 4683.3
    thus, a = 283.836
    and thats a really large number... i am doubting this.
     
    Last edited: Jun 20, 2007
  8. Jun 20, 2007 #7

    nrqed

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    I have a hard time following your reasoning. But why not simply use

    [itex] y(11s) = y_i + v_{yi} t + \frac{a_y t^2}{2} [/itex]

    Using y_i=0, v_(yi) =0, just plug in the numbers and solve for a_y! That's it. Notice that you should NOT add or subtract any 9.8 m/s^2 ! there is only one acceleration in this problem. That's it, that' sthe acceleration of the rocket!
     
  9. Jun 20, 2007 #8

    ere

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    i understand the part where there is only one accel. but how can i get the y(11s)? all i have is the position after a 4s freefall, and to start with the freefall has an initial velocity.
     
  10. Jun 20, 2007 #9

    nrqed

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    Ah, sorry. My mistake! I thought the altitude given was at 11 seconds. I will check your calculation and get back to you in a couple of minutes.
     
  11. Jun 20, 2007 #10

    nrqed

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    On the lhs it should be s(@11s), not v! (that might simply be a typo)
    But gravity is acting during the last 4 seconds, so the last term should be
    [itex]-1/2 gt^2 [/itex]

    let me do it from scratch. You had the right ideas

    During the first 11 seconds we have

    [itex]y(11 s) = \frac{a_y (11)^2}{2} = 60.5 a_y [/itex] where a_y is the acceleration of the rocket which is unknown (notice: no subtraction ro addition of g here!)

    and

    [itex] v_y(11 s) = a_y t = 11 a_y [/itex]


    for the total flight

    [itex] 4600 m = y(11 s) + v_y (11s) \times 4 seconds - \frac{ g (4 s)^2}{2} [/itex]

    Plug in the formula for y(11s) and v_y(11 s) from above and this gives you an equation for a_y to solve.


    Hope this is clear.
     
  12. Jun 20, 2007 #11

    ere

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    yep. thanks a lot!
     
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