A toy rocket is shot vertically into the air from a launching pad 9 feet above the ground with an initial velocity of 168 feet per second. The height h, in

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