You can do this problem using two different techniques for finding the vertex of a quadratic equation.
1) put the problem in Standard form of a quadratic: y=ax2+bx+c
f(x)=20x-16x2 could become f(x)=-16x2+20x. if you just switch the locations of the terms on the same side.
a=-16 b=20 and c=0
Use the formula for the vertex from Standard form: vertex at (-b/(2a), f(-b/(2a))
So the vertex's x-value will be -b/(2a) = -(20)/(2*-16)=-20/-32=5/8. super fast.
then put 5/8 in for x in the original equation f(x)=-16x2+20x to find the max height, 6.25 feet.
2) Alternatively, you could find the zeros of the function (the x-intercepts-when it hits the ground) by factoring. Then average the x values to find the time of the max height.
So solve 0=20x-16x2
factor out the GCF, 4x first: 0=4x(5-4x)
Then use the zero product property to solve 4x=0 and 5-4x=0. This will give you x-intercepts of 0 and 5/4.
The average of 0 and 5/4 is 5/8. That is the time of the max height. So again, substitute into the original equation to find the max height of 6.25 feet.