There are many ways to solve this, but if we ignore calculus and derivations from physics motions under constant acceleration, we can either find the midpoint of the two zeros of the function or we can more directly view the maximum height if we translate the quadratic into vertex form. Personally the easiest way for simple quadratics like this is to find the midpoint of the two zeros of the function...
h(t)=120t-16t^2, h(t)=0 when
16t^2-120t=0
4t(4t-30)=0 so the two zeros are when t=0 and 30/4
t=0 and 7.5
So the midpoint is 7.5/2=3.75
h(3.75)=-16t^2+120t-225 ft
Now if we did do the vertex form, which is important because it shows a general solution for all quadratics vertexes, which are the maximum/minimum points for all parabolas.
It is useful to commit to memory that the vertex, ie minimum/maximum point for all quadratics of the form ax^2+bx+c=y is:
(-b/(2a), (4ac-b^2)/(4a)) Again, this is very important as it is an absolute minimum/maximum, ie vertex for all parabolas...
In this case we are only concerned with the maximum height, or the y coordinate of the vertex, which is
(4ac-b^2)/(4a) which is in this instance (0-120^2)/(-64)=225 ft
