Question 971833
This is a dynamics/mechanics (physics) problem.
I do not know of a physics help website, but I can help.
Also, any cheerleader (and his/her fifth grader younger sibling)
would also be able to tell you that no team would be able to reach such ridiculous heights.
 
PART 1:
{{{g=32}}}{{{ft/s^2}}}= acceleration of gravity.
{{{t}}}= time to maximum height (in seconds).
{{{initial}}}{{{upwards}}}{{{velocity=20}}}{{{ft/s}}} .
During the upwards flight, {{{upwards}}}{{{velocity}}} (in ft/second) decreases linearly as
{{{upwards}}}{{{velocity=20-gt}}} ,
At the highest point of that flyer flight,
{{{t[apex]}}}= time to reach maximum height (in seconds).
and {{{upwards}}}{{{velocity=0}}} , so
so {{{average}}}{{{upwards}}}{{{velocity=(20+0)/2=10}}} ,
and at that apex, the upwards velocity in ft/s is
{{{20-gt[apex]=0}}}-->{{{20=gt[apex]}}}}-->{{{20=32t[apex]}}}-->{{{t[apex]=20/32=5/8=0.625<1}}} .
Flying at an average upwards velocity of {{{10}}}{{{ft/s}}} for {{{5.8}}} seconds,
the flyer's height would increase by less than {{{10*1ft=10ft}}}.
The increase in height can be calculated as {{{10*0.625}}}{{{ft=t6.25ft}}} .
With that height increase added to the initial {{{4ft}}} height,
the flyer's greatest attained height would be less than {{{(4+10)ft=14ft}}} ,
and actually {{{(4+6.25)ft=10.25ft}}} .
The height attained should be impressive enough (and scary enough).
Other calculation options including memorized complicated formulas are possible.
Two sentences:
Given the initial velocity of 20 ft/second,
the average velocity while going up would be 10ft/s,
and the time to maximum height would be less than 1 second.
Flying at an average velocity of 10 ft/second for less than 1 second,
the height would increase by less than 10 ft,
which added to the initial 4 ft height is far less than 25 ft.
 
PART 2 my way:
{{{v[0]}}}= initial upwards velocity, in feet/second, and
At the highest point in the flight,
the upwards velocity has gone down by {{{v[0]-0=v[0]}}}{{{ft/s}}},
and the time to achieve that downward acceleration at a rate of {{{32}}}{{{ft/s}}} per second is
{{{v[0]/32}}} seconds.
The average velocity while flying up was {{{(v[0]+0)/2=v[0]/2}}}{{{ft/s}}} .
to increase the flyers height by {{{25ft-4ft=21ft}}}
in {{{v[0]/32}}} seconds at an average velocity of {{{v[0]/2}}}{{{ft/s}}} ,
what's needed is
{{{(v[0]/32)(v[0]/2)=21}}}<-->{{{v[0]^2/64=21}}}<-->{{{v[0]^2=21*64}}}<-->{{{v[0]=sqrt(21*64)=8sqrt(21)=about36.66}}}
 
The complicated way:
Let's dust the formula that you are probably expected to memorize,
so you can see another way to do the calculations. It's
{{{height=height[0]+v[0]t-gt^2/2}}} , with 
{{{t}}}= time in the air, in seconds,
{{{height}}}= height at {{{t}}} seconds
{{{height[0]}}}= initial height, in feet
{{{v[0]}}}= initial upwards velocity, in feet/second, and
{{{g=32}}}= acceleration of gravity in {{{ft/s^2}}} .
Substituting the known values we get,
{{{height=4+v[0]t-16t^2}}} .
So {{{height}}} is a quadratic function of {{{t}}} .
All quadratic functions (generally written as y as a function of x) can be written as
{{{y=ax^2+bx+c}}} and if {{{a<0}}} ,
have a maximum at {{{x=-b/2a}}} ,
and that maximum value is {{{(4ac-b^2)/4a=c-b^2/4a}}} .
So {{{height=4+v[0]t-16t^2}}} with {{{system(a=-16, b=v[0],c=4)}}} will have a maximum for
{{{t=-v[0]/(2(-16))=v[0]/32}}} ,
and for that maximum to be {{{25}}} feet, we need
{{{4-v[0]^2/(4(-16))=25}}}<-->{{{4+v[0]^2/64=25}}}<-->{{{v[0]^2/64=25-4}}}<-->{{{v[0]^2/64=21}}}<-->{{{v[0]^2=21*64}}}<-->{{{v[0]=sqrt(21*64)=8sqrt(21)=about36.66}}}