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  1. Jan 23, 2005 #1
    I need help solving a physics problem.
    The question is a rocket is launched at an angle of 53o above the horizontal with an initial speed of 100 m/s. It moves for 3 s. along its initial line of motion with an acceleration of 30.0 m/s2. At this time its engines fail and the rocket proceeds to move as a projectile. Find (a) the max altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

    The answers in the back of the book have a being 1520, b being 36.1 and c being 4050.
    I have tried working on this problem for some time now and I get (a) being 435 so it doesnt allow me to find b and c.
    If someone could please help me out I'd really appreciate it. Thanks!
     
  2. jcsd
  3. Jan 23, 2005 #2

    dextercioby

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    At the moment the acceleration is zero (the engine fails),you must compute the velocity and its angle to the coordinate axis and the position wrt the Ox & Oy axis...From the on,it would be simple,really simple.
    What forces act on the rocket during the time the engine is running??What about afterwards??

    Daniel.
     
  4. Jan 23, 2005 #3
    I don't understand. Let me tell you what I did. I took 100 m/s and i multiplied it by 3 s. I then took that answer and added it when i took 3, squared that and mulitplied that by 30 and took half of that answer. That came out to be 435. So, to find the max height i used the equation to find the displacement. I dont see what I need to do to get 1520.
     
  5. Jan 23, 2005 #4

    dextercioby

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    That's just wrong...Plain wrong...

    [tex] \vec{r}=\vec{r}_{0}+\vec{v}_{0}t+\frac{1}{2}\vec{a}t^{2} [/tex]

    Use this equation and the geometry of the problem (plus the fact that the acceleration on the first part of the ascending trajectory is not really the one given by the engine) to solve it...

    Daniel.
     
  6. Jan 23, 2005 #5
    what does the Ro stand for? Thanks for the help too. I appreciate it
     
  7. Jan 23, 2005 #6

    dextercioby

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    Aaaa,u mean [tex] \vec{r}_{0} [/tex] ...Okay.It's the traditional notation for the position vector at the initial moment of time (chosen by me to be zero)...
    Pretty straightforward...

    What is the acceleration on the first part of the ascending trajectory??"x"-comp,"y" component...

    Daniel.
     
  8. Jan 23, 2005 #7
    I think for that it would be 79.8 and 60.2. I took the cosine and sine of 53 and multiplied it by 100
     
  9. Jan 23, 2005 #8

    dextercioby

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    Okay,those are the values for the initial velocity components.U need the initial values for the acceleration vector components..;

    Daniel.
     
  10. Jan 23, 2005 #9
    ahh ok. 23.6 and 18.05
     
  11. Jan 23, 2005 #10

    dextercioby

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    Did u forget to drop 9.8m/s^{2}from the initial "y" component??It seems that way...

    Daniel.
     
  12. Jan 23, 2005 #11
    yeah, I did thanks. Ok, now i have those numbers I dont understand how I would use those numbers to get the max height.
     
  13. Jan 23, 2005 #12

    dextercioby

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    Okay,project the vector equation i gave and use the 4 numbers to determine the other 4 numbers (2 for velocity & 2 for coordinate) of the point from where the rocket is in free fall...

    Daniel.
     
  14. Jan 23, 2005 #13
    I'm sorry but I dont think I understand. I projected the equation and I dont know what I'm trying to solve and what to plug in. I'm sorry I'm stupid about this stuff
     
  15. Jan 23, 2005 #14
    I found this posted about the same problem I'm working on

    I am solving it for a genral solution, v be the initial velocity, h be the max. height, a be the acceleration, t be the time during which it accelerated, x is angle of projection.
    h1 = vt - 0.5(g-asinx)t^2
    (g - asinx) net acceleration in downward direction.
    now it already had velocity so it raise further
    h2 = [(vf^2)/2g] where vf is the final velocity when power is off.
    h = h1 + h2
    Total time of flight T,
    This one is a shortcut tech. max. height attained h, let the body is droped from heigh h and then solve it for time taken
    t1 = (2h/a)^0.5 a is total acceleration.
    t1 is the time taken to fall, but under standerd condition time taken to attain the height h willalso be t1
    T = 2*t1

    I still dont understand what it's talking about but it might be easier for you to explain with these equations or the one you gave me. It doesnt matter either way thanks for doin this
     
  16. Jan 23, 2005 #15
    What are you having problems on?
     
  17. Jan 23, 2005 #16
    I dont understand which equation to use. I was talking to the other guy on here and he was trying to explain it to me but i didnt understand. He gave me some equation above and I dont know how to plug in my numbers into that equation.
     
  18. Jan 23, 2005 #17
    can anyone help me?
     
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