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Homework Help: Newtons law questions (rocket)

  1. Nov 25, 2009 #1
    Question reads: A rocket of mass 1000 kg is being fired to a height of 5000m. The rocket engine shuts off when the rocket reaches a height of 1000m, and the rocket coasts up to 5000m.

    a) what velocity must the rocket have at the 1000m point to enable it to reach 5000m ?
    b) what acceleraation did the rocket experience while the engine was on? and off?

    so this is a very new question for me as i was previously dealing with tension and friction forces along the plane.
    im not very sure how to start questions like this one where i need to combine kinematics.

    will i need to find the acceleration first? i know that the forces acting on the rocket is

    force applied [up] and gravitational force [ down].

    please further guide me with this problem as i have no clue as of this time.
  2. jcsd
  3. Nov 25, 2009 #2
    I think you need to find the velocity at 1000m first. Once you know that you can find the acceleration required to get it to that speed.

    First let up be 'positive' and down be 'negative' (this is arbitrary)

    a) Let it have velocity 'u' at 1000m. From that point it travels 4000m up with only the acceleration of gravity (-9.8m/s^2) to slow it down. Since its highest point is 5000m, it must have 0 velocity at that point.

    So we can use the equation [tex]v^2 = u^2 +2ax[/tex]

    With [tex]v = 0[/tex], [tex]a = -9.8[/tex] and [tex]x = 4000[/tex]

    Solving we get [tex]u = \pm 280[/tex]. But since we know it must be moving up, we have [tex]u = 280[/tex], i.e. the rocket is moving at 280m/s up. (the negative sign corresponds to when the rocket is falling back down).

    b) Before launch, u = 0 m/s, at 1000m we must have v = 280m/s, and it travels 1000m up to this point

    So we can use [tex]v^2 = u^2+2ax[/tex] again, and solving for a we get [tex]a = 39.2[/tex].

    So the acceleration when the engine is on is 39.2m/s^2 up.

    When the engine is off, only gravity is affecting the rocket, so the acceleration is 9.8m/s^2 down (or -9.8m/s^2 up).
  4. Nov 25, 2009 #3
    haha great, yea i actually ended up solving this problem 10 minutes after posting.
    my answers match the ones that you have posted.
    great, thanks ~
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