Atoy 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

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

Whose mistake? can you give more details to what the question is asking?