feet, of the rocket above the ground as t seconds after launch is given by the function h(t)=-16t^2+168t+9. How long will it take the rocket to reach its maximum height? What is the maximum height?

Y(initial) = 9

V(initial) = 168

V(final) = 0

g(accel-grav) = 32 (in feet per second

squared)

Use the following equation:

V(final) = V(initial) + a(t)

Since this object is moving straight up and

down, a = -32

Enter the knowns into the equation

0 = 168 - 32t

32t = 168

t = 168/9.8

t = 5.25

Now that you know the time it takes to

reach it’s maximum, use the general

kinematics equation to solve for final

distance:

y(final) = y(initial) + V(initial)(t) + (1/2)a

(t²)

y(final) = 9 + 168(5.25) + (1/2)(-32)

(5.25²)

= 9 + 882 - 441

= 450 feet

The maximum or minimum of a function is the point

where its first derivative = 0.

h(t) = -16t² + 168t + 9

First derivative h’(t) = -32t + 168

h’(t) = 0 = -32t + 168 = 0

Add 32t to each side: 32t = 168

Divide each side by 32: *t = 5.25 seconds*

h(t) = -16t² + 168t + 9

Height at 5.25 seconds = -16(5.25²) + 168(5.25) + 9 =

-16(27.5625) + 882 + 9 =

- 441 + 882 + 9 =* 450-ft *