Question 1198646
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I'll break the four questions into part (a) through part (d).


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Part (a)


The rocket starts on the roof at time t = 0.
Plug this value into the equation to get
h(t) = -5t^2+22t+15
h(0) = -5(0)^2+22(0)+15
h(0) = 15


The school's height is 15 meters.


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Part (b)


The rocket hits the ground at a height of h = 0.
We'll replace h(t) with 0 and apply the quadratic formula to solve.


h(t) = -5t^2+22t+15
0 = -5t^2+22t+15
-5t^2+22t+15 = 0


Plug in a = -5, b = 22, c = 15
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-22+-sqrt((22)^2-4(-5)(15)))/(2(-5))}}}


{{{t = (-22+-sqrt(784))/(-10)}}}


{{{t = (-22+-  28)/(-10)}}}


{{{t = (-22+28)/(-10)}}} or {{{t = (-22-28)/(-10)}}}


{{{t = (6)/(-10)}}} or  {{{t = (-50)/(-10)}}}


{{{t = -0.6}}} or  {{{t = 5}}}
The negative time value doesn't make much sense, so we ignore it.
t = 5 is the only practical solution here. 


At the time stamp of 5 seconds is when the rocket hits the ground.


Note: The rocket does not fall straight down or else it would hit the roof of the school, rather than the ground. 
In other words, the rocket must drift to one side as it falls.


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Part (c)


I'll replace t with x, and replace h with y.


The equation we have is y = -5x^2+22x+15
Which when compared to y = ax^2+bx+c we have
a = -5
b = 22
c = 15


Plug the 'a' and 'b' values into this formula
h = -b/(2a)
h = -22/(2(-5))
h = 2.2
This is the x coordinate of the vertex (h,k)


Then use that to find the value of the y coordinate of the vertex.
y = -5x^2+22x+15
y = -5(2.2)^2+22(2.2)+15
y = 39.2


The vertex is located at (2.2, 39.2) which is the highest point of the parabola.
The rocket reaches its max height of 39.2 meters at exactly the timestamp of 2.2 seconds.


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Part (d)


This is similar to the initial approach taken in part (b). 
This time we replace h with 10 instead of 0. The quadratic formula will show up again.


h(t) = -5t^2+22t+15
10 = -5t^2+22t+15
0 = -5t^2+22t+15-10
0 = -5t^2+22t+5


Now turn to the quadratic formula.
This time we plug in: 
a = -5
b = 22
c = 5
to get the following
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-22+-sqrt((22)^2-4(-5)(5)))/(2(-5))}}}


{{{t = (-22+-sqrt(584))/(-10)}}}


{{{t = (-22+-  24.166092)/(-10)}}}


{{{t = (-22+24.166092)/(-10)}}} or {{{t = (-22-24.166092)/(-10)}}}


{{{t = (2.166092)/(-10)}}} or  {{{t = (-46.16609)/(-10)}}}


{{{t = -0.216609}}} or  {{{t = 4.616609}}}
The decimal values are approximate.
Like earlier we ignore the negative result.


The rocket reaches a height of exactly 10 meters at the timestamp of roughly 4.616609 seconds.


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Side note: All of this ignores air resistance which greatly complicates the problem. 
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