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2 dimensional kinematics involving a rocket

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A rocket is initially at rest on the ground. At time = 0, its engines ignite causing the rocket to move in a straight line with constant total acceleration of a = 30 m/s^2 at an angle of 76 degrees, and the engines are strong enough to counter gravity and keep it exactly in a straight line. The rocket stays in a straight line with increasing speed for 15 seconds until the engines fail and the rocket goes into free fall. In this problem, the acceleration of gravity is approximated to g= 10 m/s^2. X-direction is the ground and Y-direction is along the upward direction

    a) Find the position and velocity vectors of the rocket right before the engines fail.

    2. Relevant equations
    = Vit + .5(a)(t)^2
    V = at

    3. The attempt at a solution
    When I tried to solve it, I split up the Acceleration into two components, the x and the y. Ax = 30cos(76) = 7.26 m/s^2 and Ay = 30sin(76) = 29.11 m/s^2

    Then, I plugged the acceleration components into the V= at formula respectively. So Vx = (7.26)(15) = 108.9 m/s and Vy = (29.11)(15) = 436.65

    Now that I found the velocity vectors, I proceeded to find the position vectors.
    ΔX = (0)(15) + .5(7.26)(15)^2 = 816.75 m. I set Vi as 0 because it started at rest.
    ΔY = (0)(15) + .5(29.11)(15)^2 = 3274.88 m. Same thing with the above.

    I have doubts on my answer, because the values look too huge. So I think I did something wrong in terms of either formula usage, or splitting up the acceleration into components or both.
  2. jcsd
  3. Oct 8, 2014 #2


    User Avatar

    Staff: Mentor

    Your method and results look fine.

    You could also have taken advantage of the fact that the motion was in a straight line and computed the distance along that straight line and the speed along that straight line using the usual kinematic equations, then extracted the components from them afterwards (because you know the angle of the line with respect to the horizontal).
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