SOLUTION: 4.A model rocket is shot straight up from the roof of a school. The height, h, in meters, after t seconds can be approximated by h(t)=-5t2+22t+15. Answer the following questions al

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 4.A model rocket is shot straight up from the roof of a school. The height, h, in meters, after t seconds can be approximated by h(t)=-5t2+22t+15. Answer the following questions al      Log On


   

Question 1198646: 4.A model rocket is shot straight up from the roof of a school. The height, h, in meters, after t seconds can be approximated by h(t)=-5t2+22t+15. Answer the following questions algebraically, showing all of your work.

What is the height of the school?

When does the rocket hit the ground?

What is the maximum height of the rocket?
How long does it take for the rocket to pass a window that is 10m above the ground? (hint: make h=10, group like terms, then solve the quadratic equation)]

Found 2 solutions by math_tutor2020, Alan3354:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

t+=+%28-22%2B-sqrt%28%2822%29%5E2-4%28-5%29%2815%29%29%29%2F%282%28-5%29%29

t+=+%28-22%2B-sqrt%28784%29%29%2F%28-10%29

t+=+%28-22%2B-++28%29%2F%28-10%29

t+=+%28-22%2B28%29%2F%28-10%29 or t+=+%28-22-28%29%2F%28-10%29

t+=+%286%29%2F%28-10%29 or t+=+%28-50%29%2F%28-10%29

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+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

t+=+%28-22%2B-sqrt%28%2822%29%5E2-4%28-5%29%285%29%29%29%2F%282%28-5%29%29

t+=+%28-22%2B-sqrt%28584%29%29%2F%28-10%29

t+=+%28-22%2B-++24.166092%29%2F%28-10%29

t+=+%28-22%2B24.166092%29%2F%28-10%29 or t+=+%28-22-24.166092%29%2F%28-10%29

t+=+%282.166092%29%2F%28-10%29 or t+=+%28-46.16609%29%2F%28-10%29

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.

===============================================

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

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
That's not a rocket, it's a projectile.
Rackets accelerate upward.