Marcus is shooting a rocket from a 160 foot cliff at a velocity of 48ft/sec

Question

a)Find the time it takes the rocket to hit the ground

b)what is the maximum height the rocket reaches

c)how long does it take for it to reach the maximum height

Answer 1

(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.