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Projectile motion physics

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched at an angle of 36.87º (sin=0.6 cos=0.8) with an acceleration of 30m/s² for 20s (fuel runs out). Find a) total time of flight b) horizontal range.

    2. Relevant equations


    3. The attempt at a solution
    for the first 20 seconds:
    v= 30 * 20 = 600m/s d=30 * 20² * 0.5 = 6000m
    so vertical distance is= 6000 * 0.6 = 3600m
    and horizontal distance is= 6000 * 0.8 = 4800m

    fuel runs out
    vertical speed is v= 600 * 0.6 = 360m/s
    0 = 360 - 10*t .: t=36s (time to reach max height)
    max height is= 3600 + 360*36 - 10*36²*0.5 = 10,080m

    after reaching max height
    10080 = 10 * t² * 0.5 .: t=45s (time to land)
    t=36+45=81s (amount of time gravity is doing it's job to make it fall)
    horizontal speed is v= 600*0.8 = 480m/s
    horizontal distance = 480 * 81 = 38880m

    total horizontal distance = 4800 + 38880 = 43680m
    total time of flight = 20 + 81 = 101s


    Well, it's wrong. The answer should be 41,4km and 125s.
    Help me please, thanks in advance.
     
  2. jcsd
  3. Mar 21, 2015 #2

    Bystander

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  4. Mar 21, 2015 #3

    RUber

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    125 seconds seems to match with 36.87 degrees off of vertical rather than horizontal.
    Try the calculations using 0 degrees as vertical.
     
  5. Mar 21, 2015 #4

    NascentOxygen

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    This is saying that its path for the first 20 secs will be a straight line??
     
  6. Mar 21, 2015 #5

    RUber

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    Since acceleration is given and not thrust, it sounds like this is a straight line simplified problem.
     
  7. Mar 21, 2015 #6
    Thanks everyone, I figured out what's wrong. I swaped the 0.6's with the 0.8's
    Now I'm getting the right results!
    Thanks again, see ya
     
  8. Mar 22, 2015 #7

    mfb

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    The problem statement is a bit ambiguous in terms of gravity for the first 20 seconds, but as you get the correct result you chose the intended interpretation apparently.

    Every rocket accelerating from rest will move in a straight line if the direction and magnitude of thrust does not change - constant acceleration from rest gives a straight line.
     
  9. Mar 22, 2015 #8

    NascentOxygen

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    Indeed, and for fireworks rockets that direction is fixed as "straight out the back" and they invariably follow an arc across the sky. Practically every rocket we observe launched at a low angle does not follow a straight line.

    The question could be improved, perhaps by changing to 'a guided missile', so the reader is made aware it is steerable.
     
  10. Mar 22, 2015 #9

    mfb

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    Usually because thrust or the orientation of the rocket change (e.g. firework rockets have a short acceleration phase and then fly under the influence of gravity), but here the direction and magnitude is fixed, so the rocket follows a straight line until it runs out of fuel.
     
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