Question

A ball 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?

Answer 1

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