College Algebra

Question

A rocket is shot into the air from a launching pad 15 feet above the ground with an initial velocity of 174 feet per second. The height, h, in feet of the rocket above the ground t seconds after the laugh is given by the function h(t)= -16t^2+174t+15

a) What is the maximum height of the rocket?

b) For what time interval is the rocket above 300 feet in the air ? (Round answer to the nearest tenth).

Answer 1

h'(t) = -32t + 174 = 0

32t = 174

t = 174/32 = 87/16 = 5.4375 seconds when the rocket is at maximum height

plug that into the h(t) equation to get the maximum height

h(87/16) = -16(87/16)^2 + 174(87/16) + 15 = -473.0625 + 946.125 + 15 = 488.0625 feet

h(t)>300 when 300<-16t^2 +174t +15

-16t^2 + 174t -285= 0, solve for t to find the two times when the rocket > 300 feet. then take the difference between those 2 times to get the total time above or equal to 300 feet

t = 174/32 + or - (1/32)sqr(174^2 - 4(16)(285)) = 87/16 + or - 3.4284 = 5.4375 + or - 3.4284

just double the 3.4284 to get 6.8568 seconds as the time the rocket is greater or equal to 300 feet

about 6.9 seconds

It's at that > 300 ft height from about 2 seconds to almost 9 seconds after it was first shot into the air