please help meee

Question

A rocket is launched from atop a 101 foot cliff with an initial velocity of 116 ft/s.

a. How long will it take the rocket to get to it's maximum height? 

b. What is the rocket's maximum height? 

c. When will the rocket hit the ground? 

d. At what two times will the rocket be 305 feet off of the ground?  and  (order from least to greatest)

e. For how long will the rocket be 305 feet or more off of the ground?

Answer 1

h = -16^2 + 116t + 101

h' = -32t + 116 = 0

t = 116/32 = 3.625 seconds to reach max height

max h = -16(116/32)^2 + 116(116/32) + 101 = 311.25 feet

It hits the ground when h=0 = -16t^2 +116t + 101

t = -116/-32 + (1/32)sqr(116^2+4(16)(101)) = 8,036 seconds

305 = -16t^2 + 116t + 101

-16^2 + 116t - 204 = 0

t = 3.625 + or - (1/32)sqr(116^2 -4(16)(204)) = 3 or 4.25 seconds at 305 feet

at 3 seconds it's 305 feet and going up, at 4.25 seconds it's 305 feet on the way down

time above 305 feet is 4.25-3 = 1.25 or 1 1/4 seconds