(a) Taking the top of the cliff as the reference point for the height s we have:
s=ut-½gt² where u=48ft/s and g=32 ft/s², the acceleration of gravity, and t=time.
The ground is at s=-160 ft.
-160=48t-16t², so dividing by 16 and rearranging, t²-3t-10=0=(t-5)(t+2).
So we take the positive root t=5 secs as the time taken for the rocket to fall to the ground.
(b) To find the maximum height we rewrite the equation:
s=48t-16t² as s=16(3t-t²)=-16(t²-3t+9/4-9/4)=-16(t-3/2)²+36.
When t=3/2, s is maximum at 36 feet above the hill top so this is 36+160=196 feet above the ground.
(c) Time to reach maximum height is 3/2=1.5 seconds.