# 2D motion

1. Jan 31, 2008

### clope023

1. The problem statement, all variables and given/known data

A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax= $$\alpha$$t^2 and ay=$$\beta$$-$$\gamma$$t , where $$\alpha$$=2.50m/s^4, $$\beta$$=9.00m/s^2, and $$\gamma$$=1.40m/s^3. At t=0 the rocket is at the origin and has velocity $$\overline{v0}$$ = v0x$$\widehat{i}$$ + v0y$$\widehat{j}$$with v0x = 1m/s and v0y = 7m/s.

question: what is the rockets' maximum height reached?

2. Relevant equations

1. $$\overline{v}$$(t) = v0x + $$\alpha$$t^3/3, v0y + ($$\beta$$t-$$\gamma$$t^2/2)

2. $$\overline{r}$$(t) = v0xt + $$\alpha$$t^4/12, v0yt + ($$\beta$$t^2/2 - $$\gamma$$t^3/6)

other equations I tried, might help

3. x = (v0cos$$\alpha$$0)t

4. y = (v0sin$$\alpha$$0)t-1/2gt^2

5. vx = v0cos$$\alpha$$0

6. vy = v0sin$$\alpha$$0-gt

7. y-y0 = v0yt+1/2gt^2

8. quadratic equation -b = +/- $$\sqrt{b^2-4ac}$$/2a

3. The attempt at a solution

this is a masteringphysics problem in which I was supposed to originally find the velocity and position vectors which easily enough I found to be the integral of the acceleration and the integral of velocity respectively.

I originally tried using the quadratic equation with the components of the y component of equation 2 to solve for time and plug that time back into the y component of the position vector and solve for y.

problem was I kept on getting very tiny numbers and eventually I ran out of attempts and the masteringphysics system just gave the answer to be 341m, I have no idea how they did that with the info presented, if anyone could give me any hints as to how the question is done that would be very helpful.