Aflying squirrel jumped from a tree 11 feet in the air at an initial velocity of 9 feet per sencond. the equation h=-2t^2+9t-11 models his jump where h is height in feet and t is time in senconds.
a: what was the squirrels maximum height? (find the vertex point)
b: how many senconds after he jumped was the squirrel at his highest point?
c: when did the squirrel reach the ground?

This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :)  Anyway...

h=-2t^2+9t+11

A)

dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached

dh/dt=0 only when 4t=9, t=2.25 seconds

h(2.25)=21.125 ft  (21 1/8 ft)

B)

As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.

C)

The squirrel reaches the ground when h=0...

0=-2t^2+9t+11

-2t^2-2t+11t+11=0

-2t(t+1)+11(t+1)=0

(-2t+11)(t+1)=0, since t>0 for this problem...

-2t+11=0

-2t=-11

t=5.5 seconds.

The answer is