An underground silo contains a rocket which releases a weather balloon after the rocket reaches its maximum height. The rocket then plummets back to Earth. The height of the rocket is modeled by
h(t) = -4.9t^2 + 58.8t - 50 Where h(t) is the height of the rocket in meters and t is the time in seconds.
1. How far below ground is the silo?
2. At what time is the rocket first above ground, to the nearest tenth of a second?
3. At what time does the rocket land back on Earth, to the nearest tenth of a second?
4. Relate these two times (When the rocket is first above ground and when it lands on the ground) to intervals in which the function is positive or negative.
Please answer whichever you can. Any help is very much appreciated. :)
h(t) = -4.9t^2 + 58.8t - 50 Where h(t) is the height of the rocket in meters and t is the time in seconds.
1. How far below ground is the silo?
2. At what time is the rocket first above ground, to the nearest tenth of a second?
3. At what time does the rocket land back on Earth, to the nearest tenth of a second?
4. Relate these two times (When the rocket is first above ground and when it lands on the ground) to intervals in which the function is positive or negative.
Please answer whichever you can. Any help is very much appreciated. :)
-
h(t) = - 4.9t² + 58.8t - 50
1.) The rocket is 50 m. below ground.
2.) Ground Level: h(t) = 0:
- 4.9t² + 58.8t - 50 = 0
- 4.9t² + 58.8t = 50
- 4.9(t² - 12t) = 50
t² - 12t = 50 / - 4.9
t² - 12t = - 10
t² - 12t + 36 = 36 - 10
(t - 6)² = 26
t - 6 = √26
t - 6 = ± 5.09
t = 6 ± 5.09
If t = 6 + 5.08,
t = 11.08
If t = 6 - 5.08 ,
t = 0.92
The rocket is first above ground at 0.9 sec.
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3.) The rocket lands back on Earth after about 11 secs.
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4.) Convert to vertex form, y = a(x - h)² + k, where (h, k) is the vertex:
y = - 4.9t² + 58.8t - 50
y = - 4.9(t² - 12t) - 50
y = - 4.9(t² - 12t + 36) - 50 - [- 4.9(36)]
y = - 4.9(t - 6)² - 50 - (- 176.4)
y = - 4.9(t - 6)² - 50 + 176.4
y = - 4.9(t - 6)² + 126.4
The function is positive on the interval [0, 6].
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The function is negative on the interval [6, 11.8].
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1.) The rocket is 50 m. below ground.
2.) Ground Level: h(t) = 0:
- 4.9t² + 58.8t - 50 = 0
- 4.9t² + 58.8t = 50
- 4.9(t² - 12t) = 50
t² - 12t = 50 / - 4.9
t² - 12t = - 10
t² - 12t + 36 = 36 - 10
(t - 6)² = 26
t - 6 = √26
t - 6 = ± 5.09
t = 6 ± 5.09
If t = 6 + 5.08,
t = 11.08
If t = 6 - 5.08 ,
t = 0.92
The rocket is first above ground at 0.9 sec.
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3.) The rocket lands back on Earth after about 11 secs.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯…
4.) Convert to vertex form, y = a(x - h)² + k, where (h, k) is the vertex:
y = - 4.9t² + 58.8t - 50
y = - 4.9(t² - 12t) - 50
y = - 4.9(t² - 12t + 36) - 50 - [- 4.9(36)]
y = - 4.9(t - 6)² - 50 - (- 176.4)
y = - 4.9(t - 6)² - 50 + 176.4
y = - 4.9(t - 6)² + 126.4
The function is positive on the interval [0, 6].
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The function is negative on the interval [6, 11.8].
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯…