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Homework Help: Projectile Motion Rocket Problem

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Please check my work

    A model rocket is launched vertically upwards with an initial speed of 50m/s. It accelerates with a constant upward acceleartion of 2m/s until its engine stop at an altitude of 150m.

    2.4.1) What is the maximum height reached by the rocket?
    2.4.2) How long after lift-off does the rocket reach maximum height?
    2.4.3) How long is the rocket in the air?

    2. Relevant equations

    equations of motion

    3. The attempt at a solution

    2.4.1) To get the max height I first found the final velocity of the rocket before the engine was switched off. I got this to be 55.68m/s. I then used this to find the height reached after the engine was switched off by making it the initial velocity and using the acceleration of 9.8m/s. I got this to be 158.18m. So the max height is 308.18m.

    2.4.2) By breaking the problem into two parts(one 150m and the other 158.18m)I got the times to be 2.84s and 5.68s. Adding them up gives me 8.52s to reach its max height.

    2.4.3) Multiplied the time by two(going up and coming down) to get 17.04s.

    Please could someone check my work, any help would be greatly appreciated
  2. jcsd
  3. Jan 30, 2013 #2
    The methods in part 1 and part 2 are correct (not sure about the numbers). The method for part 3 is wrong, the acceleration profiles are different on the way up and down.
  4. Jan 30, 2013 #3
    Numbers in parts 1 and 2 are correct. Ditto the comment by voko about part 3.
  5. Jan 30, 2013 #4
    Could you refer to the question number?, on the way up I used 2m/s(given), and when the engines where switched off I used 9.8m/s.
  6. Jan 30, 2013 #5


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    Everyone agrees with what you did for 2.4.1 and 2.4.2. But you cannot simply multiply the ascent time by two to get the total time in the air. It's not symmetric. Compute descent time separately.
  7. Jan 31, 2013 #6
    Thats what I did in 2.4.2, I broke the problem up and got the times to be 2.48s and 5.48s. Adding them up gave me 8.52s to reach max height. The time going up must equal the time coming down, am I right?, and so thats why I multiplied the time by 2.
  8. Jan 31, 2013 #7
    Why is that so?
  9. Feb 1, 2013 #8
    Only if the projectile is symmetric eg. parabola.
  10. Feb 1, 2013 #9
    Is the final answer 14.77s?, because I think I hit a miss with the acceleration for the second part where I used 2m/s instead of 9.8m/s coming down.
  11. Feb 1, 2013 #10
    How long does it take for the body to come from the max height down to the ground? What equation gives this time?
  12. Feb 1, 2013 #11
    It takes 6.26s to hit the ground starting from max height? I used the equation vf = vi+a*t.
  13. Feb 1, 2013 #12
    Explain how you used the equation. What are vi and vf in your case?
  14. Feb 1, 2013 #13
    When the planes engines where switched off:
    0 = 55.68+(9.8)t
    t = 5.68s

    For the remaining 150m:
    55.68 = 50+9.8t
    5.68 = 9.8t
    t = 0.58

    Adding it up gives you 6.26s.
  15. Feb 1, 2013 #14
    Okay, is the answer 7.93s to fall from max height?, I just worked it out now.
  16. Feb 1, 2013 #15
    7.93 s is correct, assuming the max height is indeed 308.18 m.
  17. Feb 1, 2013 #16
    So the total time in the air is (8.52s to go up + 7.93s to come down) = 16.45s
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