U3.L4.HW: Quadratic Functions: Question 20 Redo

Question

I was wrong last time. So let's try this differently: Put both answers first. Then explain your answers below them.

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = -4.9t² + 76t + 164. 

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after __________ seconds.

How high above sea-level does the rocket get at its peak? 

The rocket peaks at _________ meters above sea-level.

Answer 1

h(t) = -4.9t² + 76t + 164

Splashdown will occur where h(t)=0:

0 = -4.9t² + 76t + 164

t = (-76+√(76²-4(-4.9)(164)))/(2(-4.9)) ≈ 17.43

Hence, splashdown will occur at around 17.43 seconds.

The rocket's peak will be where the y-coordinate of the vertex is.

Regular Method

x = -b/2a = -76/2(-4.9) = -76/(-9.8) = 76/9.8 ≈ 7.755 seconds

Calculus Method

h(t) = -4.9t² + 76t + 164

h'(t) = -9.8t + 76

0 = -9.8t + 76

9.8t = 76

t = 76/9.8 ≈ 7.755 seconds

h(76/9.8) ≈ 458.694 meters

Hence, the rocket's peak will be about 458.694 meters above sea level.