JP launches a rocket straight upwards from a height of 20 feet above the ground with an initial upward velocity of 85 feet per second. The equation h(t) = -16t

Question

h(t) = -16t² + 85t + 20 models the height of the rocket off of the ground, t seconds after the rocket is launched.

a) Find the height of the rocket 3 seconds after it is launched. 
b) At what time will the rocket reach the maximum height? What is the maximum height of the rocket? 
c) When will the rocket reach a height of 100 feet? 
d) At what time will the rocket hit the ground? 

Answer 1

h(t) = -16t² + 85t + 20

 

a) after 3 seconds h = -16(3) + 85*3 +20 = 227 ft from the ground

 

b) at maximum height dh/dt = 0 = -32t + 85; this gives us t = 85/32 = 2.66 seconds after launch

the maximum height therefore = -16(2.66)2 +85(2.66) + 20 = 132.9 ft off the ground

 

c) 100 =  -16t² + 85t + 20 ; or 16t² - 85t +80 = 0 

 

using the quadratic formula gives two answers t = 1.22 seconds, or 4.09 seconds, this means it reaches a height of 100 ft on its way up after 1.22 seconds, and again on its way down after 4.09 seconds.

 

d) when it hits the ground h = 0 ; therefore 0 =  -16t² + 85t + 20 ; or 16t² - 85t - 20 = 0

 

using the quadratic formula gives two answers t = 5.54 seconds, or - 0.225 seconds

 

the second answer has no practical significance, so the answer is it hits the ground 5.54 seconds after launch