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Hard Projectile Motion Problem.

  1. Oct 8, 2006 #1
    Ugh. My last projectile motion problem. I just don't get one thing in this problem. The initial velocity and the acceleration. Please help me with this problem:

    A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100m/s. It moves for 3s along its initial like of motion with an acceleration of 30m/s^2. At this time its engines fail and the rocket proceeds to move as a projectile. Find a) the maximum altitude reached by the rocket, b) its total time of flight, and c) is horizontal range.

    My work so far:
    Alright, I have a basic sketch of the problem. However, I'm not quite sure on where to get started. Basically there is a linear line that extends for 3s and has a X displacement. Then the rocket drifts a parabolic motion until it reaches the ground.

    In order to calculate the displacement of the 3s, you need to find the x and y velocity components right? Here is what confuses me, how do you find the components with both an intial velocity and acceleration?

    In my last problem I only had the inital velocity, and finding the velocity components was fairly easy. However, now I have the initial velocity+ the acceleration. How am I suppose to find the components of that?

    Do I add them up and use cosine, sine? Please help me get started. Thanks a lot! :smile:
  2. jcsd
  3. Oct 8, 2006 #2


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    In pretty much every projectile motion problem you have there's a constant acceleration g in the y direction. You usually have 0 acceleration in the x direction. You then go and deal with the x and y equations of motion separately. The same applies in this problem: you can separate the problem into motion in the x direction and motion in the y direction.

    You're given the total acceleration of the rocket, surely you can break it up to x and y directions using sines and cosines?

    You should deal with the problem in 2 phases. First the motion of the rocket before t = 3s, and then the motion of the rocket after t = 3s.
  4. Oct 8, 2006 #3
    So do you add the initial velocity to the acceleration then find the x and y components of that?

    Vy= 130m/s(Sin 53)
    Vx= 130m/s(Cos 53)

    I know I'm doing something wrong.:uhh:
  5. Oct 8, 2006 #4


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    Let's examine the problem while t < 3s:
    The long way:
    x0 = 0
    y0 = 0
    vx,0 = 100 cos(53)
    vy,0 = 100 sin(53)
    ax = ?
    ay = ?

    So, just as you've probably done a million times before:
    x(t) = ?
    y(t) = ?
    vx(t) = ?
    vy(t) = ?

    Actually, you don't really need to solve for vx(t) or vy(t) from the equations above (you easily get them by solving the total velocity v = at, and breaking that into x and y components), but it makes good practice :smile:.
    Last edited: Oct 8, 2006
  6. Oct 8, 2006 #5
    Hmm.. so the initial velocity and the acceleration have two different x and y componenets?? I still don't get where this is going. After I find the x and y components for the velocity 100m/s, what do I do? Does the acceleration have x and y components too?? This is the first problem I have encountered that has something like this. If it didn't have the accleration, it would be easier. Thats what I'm confused about. Thanks.
  7. Oct 8, 2006 #6


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    The acceleration in the y-direction is gravitational, and in the x-direction it equals 30 m/s^2, until the time the rocket engine fails.
  8. Oct 8, 2006 #7


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    Yes, the acceleration does have x and y components too. Acceleration, like velocity, is a vector. You've had acceleration in your projectile motion problems before. Although then it's been the acceleration caused by gravitation, and it has only been directed in the y direction.

    The problem would "reduce" into a simple projectile motion problem if you were given the velocity and the displacement in the x and y directions at t = 3s, right? That's where we're headed: We need to figure out what those values are.
  9. Oct 8, 2006 #8


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    Unless I've misunderstood something, you are wrong. The rocket has an acceleration of 30 m/s2, into the given direction of 53 degrees above the horizontal. That already accounts for the acceleration caused by gravity.
    Last edited: Oct 8, 2006
  10. Oct 8, 2006 #9


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    You're right, I misunderstood it. Sorry. :shy:
  11. Oct 8, 2006 #10
    Ok, I've found the x and y components of the initial velocity 100m/s


    So I can put 30m/s^2 in a parallelogram, as the resultant??

    Then the x and y components of acceleration are:

    Vx= 30m/s^2(cos 53)=18.05m/s^2
    vy=30m/s^2(sin 53)=23.96m/s^2

    Now what do I do with those components?? I can't add them together or anything right?
  12. Oct 8, 2006 #11


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    You used a bit strange symbols to denote the acceleration in x and y directions (Vx and Vy, the same you used for velocities). A better choice would be Ax and Ay. Anyways, you're probably familiar with the equations:
    x = x0 + v0t + (1/2)at2
    v = v0 + at

    Can you see how to use them to get the velocities and displacements at t = 3s, the instant the rocket's engine fails?
  13. Oct 8, 2006 #12
    Oh so when you find the distance, x, you would substitue the (X component) of acceleration for Ax??




    So I just need to find t now right??
    Last edited: Oct 8, 2006
  14. Oct 8, 2006 #13
    Yea, you are right, I should use Ax and Ay for the accel components.

    For the time, how do I find it with V=Vo+at??

    Would V=100+ 30m/s^2(3s)? therefore would it be 190m/s?

    Would Vo be 100m/s?

    And the A would be 30m/s^2??
  15. Oct 8, 2006 #14


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    Well t is obviously 3s, that's the time the problem turns into a normal projectile motion problem, remember? The equations we currently have do not hold after the engine fails, as there will be no acceleration in the x direction. This is why we are currently only considering t < 3s.

    The equation you wrote above is correct. Do the same for the y direction, and for the velocity.

    EDIT: All of the equations written above are correct. I'm, however, not quite sure what you mean by V0 = 100 m/s?
  16. Oct 8, 2006 #15
    Lol, yeah the time is 3s xP I forgot about that.

    Oh, so if I do the same for Y=Yo+Vyot+.5(g)(t)^2 I will get the height of the 3 second period of time?? Therefore it would look like a triangle right?? The distance X on the bottom leg, and the height Y on the opposite leg?

    Thanks a lot, I'm starting to get this. I was just confused with the acceleration. So after the engine runs out, the problem would have a constant velocity Vxo. Ok I will try to solve the rest of the problem from here on. Thanks again to you both!! :tongue:
  17. Oct 8, 2006 #16
    In the equation V=Vo+at

    Vo would be the original/initial velocity right? So Vo=100m/s??
  18. Oct 8, 2006 #17


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    Every equation you write has to be an equation for some direction, x or y. So, Vx = Vox + ax*t and Vy = Voy + ay*t.
  19. Oct 8, 2006 #18


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    Indeed, it'll look like the triangle you described (while t < 3s, after that it'll take the paraboloidic form of "traditional" projectile motion).

    You've made one mistake: acceleration in y direction while t < 3 s is not g.
    After the engine runs out, the rocket will also have velocity in the y direction.

    Happy problemsolving :smile:.
  20. Oct 8, 2006 #19


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    Well actually, this is a case you don't have to break into components as the acceleration is parallel to velocity. So AznBoi's equation is correct.
  21. Oct 8, 2006 #20
    OH!! I get it, oops lol. The equation V=Vo+at is to determine the final velocity, or the initial velocity of the parabolic motion after the engine stops right??



    So therefore the final Velocity would be: 190m/s

    That would be the velocity Right After the engine stops. The initial velocity of the second part of the problem, the regular parabola/projectile motion part. That velocity would remain the same until it hits the ground right?
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