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Rocket accelerating up an incline

  1. Feb 9, 2014 #1
    A test rocket is launched, starting on the ground, from rest, by
    accelerating it along an incline with constant acceleration a. The incline has length
    L, and it rises at θ degrees above the horizontal. At the instant the rocket leaves the
    incline, its engines turn off and it is subject only to gravity, g≡+9.81m/s2. (Air resistance
    can be ignored). Taking the usual x-y coordinate system, with an origin at the top edge
    of the incline...

    (a) What is the position vector when the rocket is at its highest point?
    (b) What is the position vector when the rocket is on its way back down and once again at
    the same height as the top edge of the incline?

    Your symbolic answer should only
    depend on a, L, θ, g, and/or numerical factors.

    I am so lost on this problem


    My attempt:
    Height of the incline = LSinθ
    Angle: θ=arcsin(Height/L)

    Horizontal velocity after rocket has finished accelerating up the incline = cosθ * sqrt(2aL)
    Vertical velocity after the rocket has finished accelerating up the incline = sinθ * sqrt(2aL)
     
    Last edited: Feb 9, 2014
  2. jcsd
  3. Feb 9, 2014 #2

    haruspex

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    What equations do you know for constant acceleration motion? (Often called SUVAT equations.)
     
  4. Feb 9, 2014 #3
    v^2 = v0^2 + 2a(x − x0)

    x = x0 + v0t + ½at^2

    v = v0 + at
     
  5. Feb 9, 2014 #4

    haruspex

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    Right. Each SUVAT equation involves four of five variables. To pick the one to use, you see which three variables you know and what variable you are trying to find. Then look for the equation using those four.
    Now, thinking about the vertical direction, from leaving the launch ramp to maximum height, what three variables do you know and what variable are you trying to find?
     
  6. Feb 9, 2014 #5
    we need vertical velocity to be 0 to find maximum height after leaving the ramp. this means we would have to find t when velocity equals 0, but after we find t what should I do? is there another SUVAT equation that yields y position at a specific time?
     
  7. Feb 9, 2014 #6

    haruspex

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    Yes, and yes.
    you will need to find t in order to get the x co-ordinate, and having found t you can use x = x0 + v0t + ½at^2 in the vertical direction. (If you only needed the max height, and didn't need to find t, you could have used v^2 = v0^2 + 2a(x − x0) to find max height.)
     
  8. Sep 19, 2014 #7
    So I know this is an old thread, but possibly someone could help me because I have the same exact problem. I kind of understand the previous posts, yet, I am confused. In order to find the maximum height, which is not the height of the ramp, I would have to first find the value of the final velocity at the end of the ramp. This final velocity will then act as my initial velocity after it leaves the ramp and the velocity at tmax is 0. Although, to find my max height, wouldn't I also have to know the height of the ramp? I am confused, I didn't want to make a new thread, so hopefully this gets seen by someone!
     
  9. Sep 19, 2014 #8

    haruspex

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    if you mean the vertical component of the velocity, yes.
    Yes, but there's enough information given to find that.
     
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