2 dimensional kinematics involving a rocket

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SUMMARY

The discussion focuses on solving a two-dimensional kinematics problem involving a rocket that accelerates at 30 m/s² at an angle of 76 degrees for 15 seconds before entering free fall. The participant correctly decomposed the acceleration into its x and y components, calculating Ax = 7.26 m/s² and Ay = 29.11 m/s². The resulting velocity vectors were found to be Vx = 108.9 m/s and Vy = 436.65 m/s, with position vectors calculated as ΔX = 816.75 m and ΔY = 3274.88 m. The method used was validated, confirming the calculations were accurate despite initial doubts about the magnitude of the results.

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  • Understanding of two-dimensional kinematics
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  • Knowledge of kinematic equations, specifically ΔX = Vit + 0.5(a)(t)^2 and V = at
  • Basic principles of projectile motion and free fall
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Homework Statement


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.

Homework Equations


ΔX [/B]= Vit + .5(a)(t)^2
V = at

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.
 
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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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