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Homework Help: Maximum altitude of a rocket launch

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A rocket is launched at an angle of 53 degrees above the horizontal with an intial speed of 100 m/s. It moves along its initial line of motion with an accel. of 30 m/s^2 for 3 s. At this time its engines fail and the rocket proceeds to move as a free body. Find the maximum altitude reached by the rocket.

    2. Relevant equations
    Just the four basic kinematics equations. . ?

    3. The attempt at a solution
    I did:

    where is the rocket when its engines fail?
    d= vit + (1/2)at^2
    = 100 sin 53 (3) + (1/2)(30)(3^2)
    = 374.59 m

    what is the rocket's speed after the engines fail?
    vf= vi + at
    =100 + 30 (3)
    =190 m/s

    So I feel like from here I should just be able to plug into vf^2= vi^2 + 2ad and find "d," using vi= 190 m/s and vf= 0. . . .But that's not the case. Why can't I do that? And where can I go from here? I have a feeling I need to find some way to express time to reach the maximum altitude, but I have no idea how I can express that.

    ~Thank you in advance!!
  2. jcsd
  3. Sep 1, 2010 #2
    At what angle is the rocket travelling when its engines cut?
  4. Sep 1, 2010 #3
    The rocket is traveling at 53 degrees above the horizontal when its engine is cut.
  5. Sep 1, 2010 #4
    So what are the components of its velocity at the time that the engine cuts? How can you use the vertical component to calculate the time it takes to reach maximum altitude? How can that time be used to find the vertical displacement of the rocket after the engines cut?
  6. Sep 1, 2010 #5
    1. Components of initial velocity: vix= 100cos53, viy= 100sin53
    2. I honestly have no idea. :( I don't know what equation to create. . how to express time. . .
    3. . . .
  7. Sep 1, 2010 #6
    When the engine fails, it's velocity is 190 m/s at 53 degrees from horizontal.

    Then in the vertical direction, you can use the equation Vf = Vi + at, where Vf = 0, Vi = 190sin53 m/s , a = -9.8 m/s^2, and isolate t.

    Go from there.
  8. Sep 1, 2010 #7
    Oh. . .
    So viy= 190sin53, vfy= 0?

    I think I can see how I could use that to find time to fall in one of the kinematics equations, but I don't see how I could express time to reach maximum altitude.
  9. Sep 1, 2010 #8
    Remeber that after engines fails, the only acceleration on the rocket is gravity.
  10. Sep 1, 2010 #9
    Is time even relevant, then?

    Or can I just do this:

    vf^2 = vi^2 -2(9.8)d
    vf= 0 (because at apogee velocity is s 0, and apogee would be the maximum altitude)
    viy= 190sin53

    and then solve for d?
  11. Sep 1, 2010 #10
    Yes, but breaking it down helps to visualize it, no? And you asked for a way to express time, so I showed you.

    Plus, it asks for max altitude, this 'd' is only from free fall. We have to add the 'd' from rocket thrust upwards.

    Hope I helped.
    Last edited: Sep 1, 2010
  12. Sep 1, 2010 #11
    Ohh. . . .Haha the time thing was a result of Yahoo answers confusion.

    Oh, and then I add the initial distance.
    Aha! I get it!!!
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