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The path of a model rocket

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m. How long after lift off does the rocket reach the maximum height?

    from College Physics by Serway and Faughn

    2. Relevant equations

    (1) [itex]v = v_0 + at[/itex] where [itex]v[/itex] denotes velocity, [itex]v_0[/itex] denotes initial velocity, [itex]a[/itex] denotes acceleration, and [itex]t[/itex] denotes time in seconds.

    (2) [itex]x = v_0t + \frac{1}{2}at^2[/itex] where [itex]x[/itex] denotes the displacement

    (3) [itex]v^2 = v_0^2 + 2ax[/itex]

    3. The attempt at a solution

    Ok so for this problem I found 2 ways to do it. One of them gives the correct answer (in the back of the book) and the other is off by a little.

    The method that gives the correct answer:


    So let the velocity of the rocket at the point where the acceleration stops be [itex]v_1[/itex] and the velocity of the rocket at the highest point be [itex]v_2[/itex]. Let the time of the rocket's acceleration be [itex]t_1[/itex] and the time of no rocket's acceleration be [itex]t_2[/itex]. Note that at the highest point, the velocity is 0, so [itex]v_2 = 0[/itex].

    Now we find [itex]v_1[/itex]. From equation (3), [itex]v_1^2 - 50.0^2 = 2 \cdot 2.00 \cdot 150[/itex] so [itex]v_1 = 55.68[/itex] m/s.

    From equation (1), we have [itex]v_1 = 50.0 + at_1[/itex], so [itex]55.68 = 50.0 + 2.00 \cdot t_1[/itex] and [itex]t_1 = 2.84[/itex] sec.

    From equation (1) again, we have [itex]v_2 = v_1 + at_2[/itex], so [itex]-55.68 = -9.8t[/itex] and [itex]t_2 = 5.68[/itex] sec.

    The answer is [itex]\boxed{8.52}[/itex] sec.


    Ok so here's the other method that I tried but it didn't give the right answer:

    Let [itex]d_2[/itex] denote the distance that the rocket travels after it stops accelerating itself. Then, by (3), [itex]0 = 55.68^2 - 2 \cdot 9.8 \cdot x[/itex], so [itex]x = 158[/itex].

    So by (2), [itex]x = 158 = 55.68 \cdot t_2 - 4.9t_2^2[/itex].

    Solving the quadratic gives [itex]t_2 = 5.49[/itex] sec or [itex]t = 5.87[/itex] sec.

    But this doesn't give the right answer :( , help?
     
  2. jcsd
  3. Jun 5, 2012 #2
    I get t=√3100/9.8=5.681 sec.
     
  4. Jun 5, 2012 #3
    Yes :) because [itex]\sqrt{3100} = 55.68[/itex] as obtained in my first method, but why doesn't the second method work?
     
  5. Jun 5, 2012 #4


    I also wonder why 2 different values for t.
    But if we put displacement=0 then,

    0=55.68-4.9t
    ttotal=55.68/4.9=11.3632
    t=5.68163
     
    Last edited: Jun 5, 2012
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