An airplane is flying east at a constant altitude of 28,000 meters. When first seen to the east of an observer, the angle of elevation to the airplane is 71.5 degrees. After 73 seconds, the angle of elevation is 51.6 degrees. Find the speed of the airplane.

Question

An airplane is flying east at a constant altitude of 28,000 meters. When first seen to the east of an observer, the angle of elevation to the airplane is 71.5 degrees. After 73 seconds, the angle of elevation is 51.6 degrees. Find the speed of the airplane.

Answer 1

tan71.5 = Y/X = 28,000 / X.

X = 28,000 / tan71.5 = 9369 m. = d1.

X = 28,000 / tan51.6 = 22,193 m.= d2.

V = d/t = (d2-d1)/t.
V = (22,193-9369) / 73 = 176 m/s.

Answer 2

Well, if I were the airplane, I would probably be flying as fast as I can to get away from all the turbulence in the sky! But to answer your question, we can use some trigonometry.

Let's call the speed of the airplane "x" meters per second. In 73 seconds, the airplane would have traveled a horizontal distance of 73x meters.

Now, let's consider the right triangle formed by the observer, the airplane, and the ground. The height of the triangle is 28,000 meters, the horizontal distance is 73x meters, and the angle of elevation changes from 71.5 degrees to 51.6 degrees.

Using trigonometry, we can set up the following equation:

tangent(71.5) = 28000 / (73x)

Simplifying this equation, we get:

tan(71.5) = 28000 / (73x)

tan(71.5) ≈ 2.7758771

Solving for x, we get:

x ≈ 28000 / (73 * 2.7758771)

x ≈ 155.85 meters per second

So, the speed of the airplane is approximately 155.85 meters per second. And now, I'm going to fly away from all this math before my circuits get too dizzy!

Answer 3

To find the speed of the airplane, we can use trigonometry and the concept of similar triangles.

Let's denote the distance the airplane has traveled in 73 seconds as "d" (in meters).

We have two right triangles in this problem: one at the initial observation point and one at the observation point 73 seconds later.

In the initial triangle, the angle of elevation is 71.5 degrees. Let's call the height of the observer from the ground point "h" (in meters).

In the triangle after 73 seconds, the angle of elevation becomes 51.6 degrees, and the height of the observer remains the same, which is "h."

Now, let's find the distance the observer has traveled horizontally in the 73-second time frame. Denote this distance as "x" (in meters).

Using trigonometry, we can write the following equation for the initial triangle:
tan(71.5 degrees) = h / x

Similarly, for the triangle after 73 seconds:
tan(51.6 degrees) = h / (x + d)

Rearrange the equations to solve for h:
h = x * tan(71.5 degrees) ---(1)
h = (x + d) * tan(51.6 degrees) ---(2)

Since h is the same for both equations, we can equate them:
x * tan(71.5 degrees) = (x + d) * tan(51.6 degrees)

Now, solve this equation for x:
x * tan(71.5 degrees) = x * tan(51.6 degrees) + d * tan(51.6 degrees)

Rearrange the equation to isolate d:
d * tan(51.6 degrees) = x * tan(71.5 degrees) - x * tan(51.6 degrees)

Simplify the equation:
d = x * (tan(71.5 degrees) - tan(51.6 degrees)) / tan(51.6 degrees)

Now, substitute the known values:
d = x * (tan(71.5 degrees) - tan(51.6 degrees)) / tan(51.6 degrees)

The known values are:
tan(71.5 degrees) ≈ 2.8699
tan(51.6 degrees) ≈ 1.2707

Substituting these values, we get:
d = x * (2.8699 - 1.2707) / 1.2707

Simplify the equation:
d = x * 1.5875

Now, we have the distance traveled by the plane in 73 seconds. To find the speed, we divide this distance by the time interval:

Speed = Distance / Time
Speed = d / 73

Substituting the value of d, we have:
Speed = (x * 1.5875) / 73

Now, we have the speed of the airplane, in meters per second.

Answer 4

To find the speed of the airplane, we need to use the information given and apply trigonometry concepts.

Let's break down the problem:

1. The airplane is flying at a constant altitude of 28,000 meters.
2. When first seen to the east of an observer, the angle of elevation to the airplane is 71.5 degrees.
3. After 73 seconds, the angle of elevation is 51.6 degrees.

We can set up a right triangle with the observer at one vertex, the airplane at another vertex, and the highest point of the airplane's flight path as the third vertex.

Now, to find the speed of the airplane, we need to find the horizontal distance it travels during the 73 seconds.

Using trigonometry, we can relate the angles of elevation to the sides of the right triangle formed by the observer, the airplane, and the highest point of the flight path:

1. Let x be the horizontal distance traveled by the airplane in meters (this is what we want to find).
2. The altitude of the airplane, 28,000 meters, is the opposite side of the 71.5 degree angle.
3. The altitude of the airplane minus the highest point it has reached is the opposite side of the 51.6 degree angle.

Using the tangent function, we can write two equations:

For the initial angle of elevation:
tan(71.5) = 28000 / x

For the final angle of elevation:
tan(51.6) = (28000 - highest point) / x

Since there is no information about the height of the highest point, we can assume it to be zero. Therefore, the second equation becomes:
tan(51.6) = 28000 / x

Now we have two equations:
tan(71.5) = 28000 / x
tan(51.6) = 28000 / x

We can solve these equations simultaneously to find the value of x, which represents the horizontal distance traveled by the airplane.

Solving the equations is beyond the scope of this explanation, but you can use a scientific calculator or mathematical software to obtain the value of x.

Once you have the value of x, you can divide it by the time of 73 seconds to find the speed of the airplane. Remember to convert the units if necessary to get the speed in meters per second.