physics

A)A model rocket is launched straight upward with an initial speed of 47.7 m/s. It accelerates with a constant upward acceleration of 2.02 m/s2 until its engines stop at an altitude of 157 m. What is the maximum height reached by the rocket?
B)How long after lift off does the rocket reach its maximum height?
C)How long is the rocket in the air?

I solved part A but I am having troubles with part B and C, Help?

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asked by Justin
  1. v = 47.7 + 2.02 t
    157 = 47.7 t + 1.01 t^2
    1.01 t^2 + 47.7 t - 157 = 0
    t = [ -47.7 +/- sqrt(2275+634) ]/2.02
    t = [ -47.7 +/- 53.9 ] / 2.02
    t = 6.24 seconds then engine stops at 157 m
    still coasting up
    v = 47.7 + 2.02(6.24) = 60.3 m/s at engine stop
    when does v = 0 at top?
    0 = 60.3 - g t
    t = 60.3/9.81 = 6.15 seconds more coasting up so we have been going 12.4 seconds (that is part B by the way)
    h = 157 + av speed * 6.15
    h = 157 + 30.15*6.15 = 342 meters at the top after 12.4 seconds aloft
    that is Part A
    now we fall from 342 meters
    342 = 4.9 t^2
    t = 8.36 second fall
    total t = 12.4 + 8.36 = 20.8 seconds aloft

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    posted by Damon
  2. during the upward travel,
    h = 47.7t + 1/2 (2.02)t^2
    solve for t when h=157
    t = 3.09 seconds.
    At that time, v = 47.7 + 3.09(2.02) = 53.94 m/s

    Now, from then on the height is given by

    h = 53.94(t-3.09) - 4.9(t-3.09)^2
    Now just find the vertex of the parabola, and solve for t when h=0.

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    posted by Steve
  3. Hmmm. If Damon disagrees, better go with his numbers, though I'd double-check to be sure.

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    posted by Steve
  4. Use Steve's result for first part. I forgot to divide by 2.o2

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    posted by Damon
  5. How do I find the vertex?

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    posted by Sam
  6. recall your algebra I.

    For ax^2+bx+c=0, the vertex is at x = -b/2a.

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    posted by Steve
  7. v = 47.7 + 2.02 t
    157 = 47.7 t + 1.01 t^2
    1.01 t^2 + 47.7 t - 157 = 0
    t = [ -47.7 +/- sqrt(2275+634) ]/2.02
    t = [ -47.7 +/- 53.9 ] / 2.02
    t = 3.07 seconds then engine stops at 157 m
    still coasting up
    v = 47.7 + 2.02(3.07) = 53.9 m/s at engine stop
    when does v = 0 at top?
    0 = 53.9 - g t
    t = 53.9/9.81 = 5.49 seconds more coasting up so we have been going 8.56 seconds (that is part B by the way)
    h = 157 + av speed * 5.49
    h = 157 + (53.9/2)(5.49) = 305 meters at the top after 8.56 seconds aloft
    that is Part A
    now we fall from 305 meters
    305 = 4.9 t^2
    t = 7.89 second fall
    total t = 8.56 + 7.89 = 16.4seconds aloft

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    posted by Damon

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