Question 1068621
The function {{{h(t)=60t-16t^2}}} , which could be written as {{{h(t)=-16t^2+60t}}} ,
is a quadratic function
(a polynomial with degree 2),
and graphs as a parabola.
Your teacher probably wants to remember all the names and formulas.
 
WITH FORMULAS:
You have probably studied quadratic functions of the form
{{{f(x)=ax^2+bx+c}}} and were told that they have
a maximum if {{{a<0}}} , or a minimum is {{{a>0}}}
at {{{x=-b/"2a"}}} .
That point in the graph is also called the vertex of the parabola.
In this case, the variable is called {{{t}}} instead of {{{x}}} ,
and the coefficients are
{{{a=-16}}} , {{{b=60}}} , and {{{c=0}}} .
So the maximum happens at {{{t=-b/"2a"=-60/(2*(-16))=-60/(-32)=15/8=1.875}}}
So the maximum is
{{{h(15/8)=-16*(15/8)^2+60*(15/8)=highlight(56.25)}}} .
 
WITH EASE (AND THINKING):
Those parabolas are symmetrical,
so if a parabola has two zeros,
the maximum or minimum happens halfway between the zeros.
This parabola must have two zeros
and a maximum in between because the ball
was thrown up from the ground (how did they do that?)
gets to some maximum height,
and falls back to the ground on the ground.
Let me find the zeros by factoring.
{{{60t-16t^2=0}}} 
{{{t(60-16t)=0}}}
So, the zeros are {{{t=0}}} and
(from {{{60-16t=0}}}<-->{{{60=16t}}}<-->{{{t=60/16}}} ) {{{t=15/4}}} .
The maximum happens halfway between {{{t=0}}} and {{{t=15/4}}} .
So, it happens at {{{t=(1/2)(15/4)=15/8}}} .
The maximum is {{{h(15/8)}}}
{{{h(t)=60t-16t^2=t(60-16t)}}}
{{{h(15/8)=(15/8)(60-16(15/8))=(15/8)(60-30)=(15/8)(30)=15*15/4=225/4=highlight(56.25)}}} .
 
Either way, the maximum height is {{{highlight(56.25feet)}}} .