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What maximum altitude does the rocket reach

  1. Oct 6, 2003 #1
    A rocket rises vertically, from rest with an acceleration of 3.2 m/s until it runs out of fuel at an altitude of 1200 m. After this point, its acceleration is that of gravity, downward. (a) what is the velocity when it runs out of fuel.(b) How long does it take to reach this point. (c) what maximum altitude does the rocket reach.(d) how much time does it take to reach maximum altitude. (e) With what velocity does it reach the earth. How long (total) is it in the air

    Thank You for you help
  2. jcsd
  3. Oct 8, 2003 #2
    this is my first post on this forum, so hello everyone :smile:

    i take it you mean that the acceleration is constant until the fuel is burnt out.

    assuming accleration is constant, we know that up to the point of fuel burn out:

    initial velocity u = 0m/s
    acceleration a = 3.2m/s
    displacement d = 1200m

    start by answering part (b):

    since acceleration is constant we can apply the equation:

    d = ut + 1/2at^2

    where t = time taken to reach 1200m. substitute the above values in to give an equation in t:

    1200 = 0t + 1/2(3.2)t^2
    1200 = 1.6t^2

    rearange to get t:

    t = √1200/1.6)
    t = √750
    t ≈ 27.386s

    now do part (a):

    a is constant so we can use:

    v = u + at

    where v is the velocity at 1200m. substitute in above values:

    v = 0 + 3.2√750
    v ≈ 87.636m/s

    part (c)

    consider the flight of the rocket after it runs out of fuel. we know:

    acceleration a ≈ -9.8m/s/s
    initial velocity u = 3.2√750m/s
    velocity at maximum altitude v= 0m/s

    constant acceleration (assuming no air resistance) so use:

    v^2 = u^2 + 2ad

    where d is the displacement of the rocket after it runs out of fuel. substitute in values and solve for d:

    0^2 = (3.2√750)^2 + 2(-9.8)d
    d = (0 - (3.2√750)^2)/2(-9.8)
    d = (-(3.2√750)^2)/-19.6
    d = (-10.24(750))/-19.6
    d ≈ 391.837m

    add this displacement to the altitude of the rocket when its fuel ran out:

    maximum altitude = 1200 + 391.837 = 1591.837m

    part (d)

    consider the flight of the rocket after it runs out of fuel. we know:

    a ≈ -9.8m/s/s
    u = 3.2√750m/s
    v = 0m/s

    use equation:

    v = u + at

    0 = 3.2√750 + (-9.8)t
    t = (-3.2√750)/-9.8
    t ≈ 8.942s

    add this to the time taken to run out of fuel:

    time taken to reach max altitude = 27.386 + 8.492 = 36.382s

    part (e)

    u = 0m/s
    a ≈ 9.8m/s/s
    d = 1591.837m
    v = velocity at 0 altitude.

    v^2 = u^2 + 2ad
    v = √(u^2 + 2ad)
    v = √31200.005
    v ≈ 176.635m/s

    total time in the air = 36.382 + 176.635/9.8 ≈ 54.406s

    all of these calculations assume constant acceleration and zero air resistance.
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