- #1

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## Homework Statement

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

question: what is the rockets' maximum height reached?

## Homework Equations

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

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

other equations I tried, might help

3. x = (v0cos[tex]\alpha[/tex]0)t

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

5. vx = v0cos[tex]\alpha[/tex]0

6. vy = v0sin[tex]\alpha[/tex]0-gt

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

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

## 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.