Question 1009579
{{{h(t)=-16t^2+v[0]*t+h[0]}}}
represents the height (in feet) after {{{t}}} seconds
of an object launched form an initial height {{{h[0]}}} feet,
with an initial upwards velocity {{{v[0]}}} ft per second.
Since you are launching those fireworks "from the ground",
{{{h[0]=0}}} for both fireworks.
 
For the first firework, {{{v[0]=70}}} , so its height {{{t}}} seconds after launch is
{{{h(t)=-16t^2+70t}}} .
That is a quadratic function with a {{{-16}}} negative coefficient for the independent variable {{{t}}} .
We know that a quadratic function {{{y=f(x)=ax^2+bx+c}}} with {{{a<0}}}
has a maximum for {{{x=-2/"2 a"}}} ,
so {{{h(t)=-16t^2+70t}}} has a maximum for
{{{t=-70/(2(-16))=(-70)/(-32)=35/16=2&1/16}}} .
That means the first firework will reach its maximum height {{{35/16=2&1/16}}} seconds after launch.
At that point its maximum height in feet will be
{{{h(35/16)=-16(35/16)^2+70(35/16)=-35^2/16+70*35/16=35^2/16=76.5625}}} .
After that, it will start to come down,
and will explode at {{{3}}} seconds, when its height in feet will be
{{{h(3)=-16(3)^2+70*3=-16*9+210=-144+210=66}}} .
 
For the second firework, {{{v[0]=90}}} , so its height in feet {{{t}}} seconds after launch is
{{{h(t)=-16t^2+90t}}} .
If it ever reaches a height of {{{130}}} ft,
that would happen at {{{t}}} seconds after its launch, with
{{{130=-16t^2+90t}}} .
Unfortunately, that second firework never reaches {{{130}}} ft.
The equation {{{130=-16t^2+90t}}}<-->{{{16t^2-90t+130=0}}} has no solution.
The second firework will reach its maximum height at
{{{t=-90/(2(-16))=(-90)/(-32)=45/16=2&13/16=2.8125}}} seconds after its launch.
At that point its maximum height in feet will be
{{{h(45/16)=-16(45/16)^2+90(45/16)=-45^2/16+90*45/16=45^2/16=76.5625}}} .
After that, it will start to come down.
At {{{3}}} seconds after its launch, the second firework's height in feet will be
{{{h(3)=-16(3)^2+90*3=-16*9+270=-144+270=126}}} .
 
Ideally, you would design the fireworks so that they would explode at their maximum height.
For the first firework, that would be {{{45/16=2&13/16=2.1875}}} seconds after its launch,
when it would be at a height of {{{35^2/16=76.5625}}} feet.
For the second firework, that would be {{{45/16=2&13/16=2.8125}}} seconds after its launch,
when it would be at a height of {{{45^2/16=126.5625}}} feet.
 
Maybe you cannot time the explosions that precisely.
For the second firework, you could design it to explode about {{{3}}} seconds after launch.
With a {{{t="3.0000"}}} second delay between launch and explosion,
it would explode at {{{126}}} feet.
That is not {{{130}}} feet, or even the maximum height of {{{126.5625}}} feet,
but it is very close to the maximum height.
For the first firework, you could design it to explode about {{{2}}} seconds after launch.
With a {{{t="2.0000"}}} second delay between launch and explosion, it would explode at
{{{h(2)=-16(2)^2+70*2=-16*4+140=-64+140=76.000}}} feet.
That is not the maximum height of {{{46.5625}}} feet,
but it is very close to the maximum height.