NASA launches a rocket at
t
=
0
seconds. Its height, in meters above sea-level, as a function of time is given by
h
(
t
)
=
−
4.9
t
2
+
166
t
+
332
.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after
seconds. (Round your answer to 2 decimals.)
How high above sea-level does the rocket get at its peak?
The rocket peaks at
meters above sea-level. (Round your answer to 2 decimals.)
S
e
e
h
o
w
a
n
n
o
y
i
n
g
v
e
r
t
i
c
a
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t
e
x
t
i
s
?
To find the time of splashdown, we need to set the height function equal to zero and solve for t.
The equation is: h(t) = -4.9t^2 + 166t + 332
Set h(t) = 0: -4.9t^2 + 166t + 332 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
For this equation, a = -4.9, b = 166, and c = 332. Substituting those values into the formula gives:
t = (-166 ± sqrt(166^2 - 4(-4.9)(332))) / (2(-4.9))
Calculating the expression inside the square root:
sqrt(166^2 - 4(-4.9)(332)) = sqrt(27556 + 6479.2) = sqrt(34035.2) ≈ 184.51
Now we have:
t = (-166 ± 184.51) / (-9.8)
Simplifying further:
t1 = (-166 + 184.51) / (-9.8) ≈ 1.84 seconds
t2 = (-166 - 184.51) / (-9.8) ≈ -3.89 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the splashdown occurs approximately 1.84 seconds after the rocket launch.
To find the height at which the rocket peaks, we can find the maximum of the height function. Since the coefficient of the t^2 term is negative (-4.9), the graph of the function is a downward facing parabola, and the maximum occurs at the vertex.
The formula for the t-coordinate of the vertex is:
t = -b / (2a)
Substituting the values, we get:
t = -166 / (2(-4.9)) ≈ 16.94 seconds
Next, we substitute this t value into the height function to find the height at this time:
h(16.94) = -4.9(16.94)^2 + 166(16.94) + 332 ≈ 1678.37 meters
Therefore, the rocket peaks at approximately 1678.37 meters above sea-level.
To find the time at which splashdown occurs, we need to find the solution to the equation h(t) = 0, where h(t) is the height of the rocket above sea-level.
Given the formula for the height function:
h(t) = -4.9t^2 + 166t + 332
Setting h(t) = 0:
-4.9t^2 + 166t + 332 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case:
a = -4.9
b = 166
c = 332
Plugging in the values into the quadratic formula:
t = (-166 ± sqrt(166^2 - 4(-4.9)(332))) / (2(-4.9))
Simplifying further:
t = (-166 ± sqrt(27556 + 6473.6)) / (-9.8)
t = (-166 ± sqrt(34029.6)) / (-9.8)
Using a calculator, we find that:
t ≈ (-166 ± 184.5) / (-9.8)
Solving for the positive value of t:
t ≈ (18.5) / (-9.8)
t ≈ -1.88 seconds
Since time cannot be negative in this context, we can ignore the negative value. Therefore, the rocket splashes down after approximately 1.88 seconds.
To find the height of the rocket at its peak, we can use the vertex formula:
t = -b / (2a)
Using the values for a and b from before:
t = -166 / (2(-4.9))
Simplifying further:
t = -166 / (-9.8)
t ≈ 16.94 seconds
Plugging this value into the height function formula:
h(16.94) = -4.9(16.94)^2 + 166(16.94) + 332
Calculating this expression, we find that the rocket peaks at approximately 1782.89 meters above sea-level.
Therefore, the answers to the questions are:
1. The rocket splashes down after approximately 1.88 seconds.
2. The rocket peaks at approximately 1782.89 meters above sea-level.