Answer:
D=90.3m.
Explanation:
Here we got a parabolic motion problem, On the Y axis we have a free-fall motion, so the time to reach the ground is given by:
[tex]y=yo+Vo*t+\frac{1}{2}*a*t^2\\0=30m+0*t-\frac{1}{2}*(-9.8)*t^2\\t=2.5s[/tex]
Having that time we now can calculate the horizontal distance traveled by the rocket.
The acceleration is a continous function, if we integrate the acceleration we obtain the velocity:
[tex]vf=\int\limits^{t}_0 {11.60*2t} \, dt\\ vf=11.6m/s^3*t^2[/tex]
[tex]Vx(t)=12.0m/s+11.6m/s^3*t^2[/tex]
the displacement is given by:
[tex]D=\int\limits^t_0 {Vx} \, dx[/tex]
[tex]D=12.0m/s*t+3.86m/s^3*t^3\\[/tex]
Substituting the time:
D=90.3 m
