Its height(in meters),x seconds after takeoff, is modeled by h(x)= -(x-11)(x+3) how many seconds after takeoff will the hovercra
ft land on the ground
2 answers:
Answer:
14s
Step-by-step explanation:
1. Since the point in this question is considering the moment the hovercraft has took off, h=0 till it lands on the ground. We can do this algebraic calculation. To find the roots of the quadratic equation.
Looking at the equation:
Also, these points mark the initial and ending point of the hovercraft path. As 11 and -3 are the roots, the time between those points is the elapsed time by the hovercraft. Considering also, the x-axis, the time.
2. Graphically, we can see that movement of the hovercraft, since a< 0. The concavity is downward.
As 11 and -3 are the roots, the time between the those points, is the elapsed time by the hovercraft
Answer:
Step-by-step explanation:
The instants when hovercraft lands on the ground are those when h(x) = 0. Given that height function is a second-order polynomial in its factorized form, such instants are related to the roots of the function.
Time is a positive real number and the only possible solution:
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I hope that's what you meant. Let me know if I'm wrong.
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So now the original question
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Answer:
1. The distance of the ship from the base of the lighthouse is approximately 1.24 miles
2. a)The horizontal distance the plane must start descending is approximately 190.81 km
b) The angle the plane's path will make with the horizontal is approximately 18.835°
3. The depth of the submarine is approximately 107.51 m
Step-by-step explanation:
The
1. From the question, we have;
The height of the observer above the water = 80 ft.
The angle of depression of the ship from the observer, θ = 0.7°
Let the position of the observer be 'O', let the location of the ship be 'S', let the point directly above the ship at the level of the observer be 'H', we have;
HS = The height of the observer = 80 ft.
Therefore, we get;
The distance of the ship from the base of the lighthouse ≈ 6,547.763 ft. ≈ 1.24 miles
2. The elevation of the plane, h = 10 km
The angle of the planes path with the ground, θ = 3°
Similar to question (1) above, the horizontal distance the plane must start descending, d = t/(tan(θ))
∴ d = 10 km/(tan(3°)) ≈ 190.81 km
The horizontal distance the plane must start descending, d = 190.81 km
b) If the pilot start descending 300 km from the airport, the angle the plane's path will make with the horizontal, θ, will be given as follows;
From trigonometry, we have;
Where the opposite leg length = The elevation of the plane = 10 km
The adjacent leg length = The horizontal distance from the airport = 300 km
The angle the plane's path will make with the horizontal, θ ≈ 18.835°
3. The angle at which the submarine makes the deep dive, θ = 21°
The distance the submarine travels along the inclined downward path, R = 300 m
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The depth, of the submarine, 'd' is given as follows;
∴ d = R × sin(θ)
d = 300 m × sin(21°) ≈ 107.51 m
The depth of the submarine ≈ 107.51 m