a.) Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? b.) How high above sea-level does the rocket get at its peak?

Question

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) = − 4.9t^2 + 220t + 262.

Answer 1

a) The rock splashes down when h(t) = 0. Set h(t) = 0 and solve for t using quadratic formula.

0 = − 4.9t^2 + 220t + 262

t = -1.16, 46.05

(Throw away the negative answer, since time can't be negative.)

t = 46.05 sec

(b) The rock is at it's peak when the slope of it's trajectory is zero. This trajectory is also the tangent line, so finding when the slope of the tangent line is zero is the goal. To do this, take the derivative of h(t) and set h'(t) = 0.

h'(t) = -9.8t + 220

0 = -9.8t + 220

t = 22.45 sec

Use this time to find the height of the ball h(t), so h(22.45 sec).

h(t) = − 4.9(22.45)^2 + 334(22.45) + 255

h(t) = 2731.39 m

Answer 2

h=0 = -4.9t^2 +220t +262

use the quadratic formula

t = 220/9.8 + (1/9.8)sqr(220^2 +4(4.9)(262))

t = 46.06 seconds to splashdown

take the derivative and set =0

h' =-9.8tt +220=0

t = 220/9.8 = 22.45 seconds to reach maximum height

plug 220/9.8 into h(t)

h(220/9.8) = -4.9(220/9.8)^2 +220(220/9.8) +262

= -48400/2(9.8) + 48400/9.8 +262

= 48400/19.6 +262

= 2,731.39 meters= max height