Physics

A rocket moves upward, starting from rest with an acceleration of 35.0 m/s2 for 5.63 s. It runs out of fuel at the end of the 5.63 s but does not stop. How high does it rise above the ground?

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asked by Monty
  1. height = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time)

    1. find height 1
    starting from rest (initial velocity=0) with an acceleration of 35.0 m/s2 (=acceleration) for 5.63 s (=time)

    2. find final velocity at height 1
    (final velocity)^2 = (initial velocity)^2 + 2*(acceleration)*(distance)

    3. find height 2
    It runs out of fuel at the end of the 5.63 s (time) but does not stop (acceleration=-9.8m/s^2 (gravity) and initial velocity of height 2 is final velocity of height 1)

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  2. 4. finally, add height 1 + height 2

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  3. Similar to the others.

    A rocket moves upward, starting from rest with an acceleration of 28.8 m/s2 for 4.13 s. It runs out of fuel at the end of the 4.13 s but does not stop. How high does it rise above the ground?

    Liftoff to burnout in 4.13 sec. Burnout velocity Vbo = 28.8(4.13) = 123.84m/s. Burnout altitude h1 = at^2/2 - gt^2/2 = 19(4.13)^2/2 = 162m. From Vbo to V = 0, Vf = Vbo - gt = 0 = 123.84 - 9.8t or t = 12.636 sec. From Vbo to V = 0, h2 = 123.84(12.636) - 9.8(12.636)^2/2 = 782m. Total height H = h1 + h2.

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