Question 1115426
This is a classical physics problem, and classical physics should be intuitive, because we see it in action in the world around us.
Alternatively, we can memorize "formulas" for every "type" of problem without really understanding.
 
THE SIMPLE/FIFTH-GRADER SOLUTION:
Somewhere in the physics book, it is stated that the acceleration of gravity increases the downward speed of an object by 32 feet per second every second:
{{{g=32}}}{{{ft/s^2}}} (with s as the symbol for second).
So, it will take {{{82/32=2.5625}}} seconds for gravity to get that initial upwards velocity down to zero, before the flare remain star having increasing downwards velocity.
During those initial {{{2.5625}}} seconds the upwards velocity decreased steadily/linearly from {{{"82 ft / s"}}} to {{{"0 ft / s"}}} .
{{{drawing(300,300,-1,4,-10,90,grid(0),
line(0,82,2.5625,0),locate(2.5,7,time),
locate(3,7,"( seconds )"),locate(0.1,88,upwards),
locate(1.15,88,velocity),locate(0.5,83,"( ft / s )")
)}}} .
During that time the average upwards velocity (in ft/s) was
{{{(82+0)/2=41}}} ,
and the upwards distance covered (in feet) was
{{{41*2.5625=105.0625}}} .
We can round that to 1 decimal place to get {{{105.1ft}}} .
As the flare started from {{{88ft}}} above ground,
the maximum height above the ground this flare will go (in feet) is
{{{88+105.1=highlight(193.1)}}} .
 
WITH FORMULAS (maybe your teacher wants that):
 
The basic formulas for an object moving in one direction with an "initial velocity" {{{v[0]}}} and a constant acceleration {{{a}}} are
{{{v=v[0]+at}}} and
{{{d=v[0]t+(1/2)at^2}}} or {{{v[avg]=(v+v[0])/2}}} (pick one).
In those formulas {{{d}}} is distance covered,
{{{t}}} is time since a time, {{{t=0}}} ,
that we can conveniently say it is the start of the observation,
an "initial velocity" is the velocity at the start of the observation,
when the velocity was {{{v[0]}}} .
From those two formulas, and your algebra skills,
you can get any other formula you will be given.
 
In particular, for projectiles shot upwards, from somewhere om planet Earth,
the height {{{h}}} above the ground is
{{{h=h[0]+v{0]t-16t^2}}} ,
where {{{h[0]}}} is the initial height above the ground, and {{{t}}} is the time since "launch".
plugging into that formula the numbers from the problem,
{{{h=88+82t-16t^2}}} .
In algebra class, you would probably write it as
{{{h=-16t^2+82t+88}}} ,
because we like to order polynomial terms from highest to lowest exponent.
In algebra 2 courses is the US, you are likely to be told that
a quadratic function {{{f(x)=ax^2+bx+c}}} has a maximum (if {{{a<0}}} )
at {{{x=(-b)/"2 a"}}} . If {{{a>0}}} it would be a minimum)
(another formula, but you do not need to memorize it if your algebra skills include "completing the square").
{{{h=-16t^2+82t+88}}} is a quadratic function of {{{t}}} (instead of {{{x}}} ),with {{{a=-16}}} and {{{b=82}}} ,
so it has a maximum at {{{t=(-b)/"2 a"=(-82)/(2*(-16))=(-82)/(-32)=5.5625}}} ,
and for that value of {{{t}}} , in seconds, the height {{{h}}} , in feet, is
{{{h=-16*2.5625^2+82*2.5625+88=-106.0625+210.125+88=193.0625}}} .
Rounding to 1 decimal place, {{{highlight(h=193.1)}}} .