Before the engines fail, the rocket's altitude at time t is given by
[tex]y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
and its velocity is
[tex]v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
The rocket then reaches an altitude of 1150 m at time t such that
[tex]1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
Solve for t to find this time to be
[tex]t=11.2\,\mathrm s[/tex]
At this time, the rocket attains a velocity of
[tex]v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}[/tex]
When it's in freefall, the rocket's altitude is given by
[tex]y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity, and its velocity is
[tex]v_2(t)=124\dfrac{\rm m}{\rm s}-gt[/tex]
(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for [tex]y_2(t)[/tex] to reach 0:
[tex]1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s[/tex]
So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.
(b) Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]
where [tex]v_f[/tex] and [tex]v_i[/tex] denote final and initial velocities, respecitively, [tex]a[/tex] denotes acceleration, and [tex]\Delta y[/tex] the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means [tex]y_2[/tex] will contain the information we need to find the maximum height.
[tex]-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)[/tex]
Solve for [tex]y_{\rm max}[/tex] and we find that the rocket reaches a maximum altitude of about 1930 m.
(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to [tex]y_2(t)[/tex]) to be about 32.6 s. Plug this into [tex]v_2(t)[/tex] to find the velocity before it crashes:
[tex]v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}[/tex]
That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.
