Aball is thrown into the air with an upward velocity of 36 ft/s. its height h in feet after t seconds is given by the function h = −16t2 + 36t + 5. in how many seconds does the ball reach its maximum height? round to the nearest hundredth if necessary. what is the ball’s maximum height?

This can be solved in two ways: With heavy tools or with just algebra. 
What is your level? Have you studied calculus? 

With pure algebra: 
We need to find the maximum of the function h = −16t^2 + 36t + 5
Lets take out -1 for simplicity: 
h = −(16t2 - 36t - 5) 
For now lets just work with this: 16t^2 - 36t - 5
16t^2=(4t)^2 
(4t-x)^2= 16t^2-2*4xt+x^2
we have -36t so x should be 4.5 as 2*4*4.5=36

Lets see what we have now:
16t^2 - 36t - 5= (4t-4.5)^2 is this true? No but close
(4t-4.5)^2= 16t^2- 2*4*4.5t +4.5^2= 16t^2-36t+20.25

16t^2 - 36t - 5   and 16t^2-36t+20.25 nearl the same just take away 25.25 from the right hand side

Getting long, just stay with me: 
16t^2 - 36t - 5= (4t-4.5)^2 - 25.25

h= -{(4t-4.5)^2 -25.25}
h=-(4t-4.5)^2 + 25.25

We want to find the maximum of this function. -(4t-4.5)^2 this bit is always negative or 0, so it maximum is when it is 0. Solve: 4t-4.5=0
t=1,125

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good