The angle of elevation of a jet plane from a point A on the ground 60 degree.After a flight of 15 seconds the angle of elevation changes to 30 degrees.If the jet plane is flying at a constant height of 1500√3m,find the speed of the jet plane?

Question

Answer

Answer

Problem:

The angle of elevation of a jet plane from a point A on the ground 60 degree.After a flight of 15 seconds the angle of elevation changes to 30 degrees.If the jet plane is flying at a constant height of 1500√3m,find the speed of the jet plane?

Solution:

Let's consider the given problem step-by-step:

Step 1: Draw a diagram

Draw a diagram to represent the given situation. Label the relevant parts with the given values and the unknowns.

![jet_plane_diagram.png](https://edurev-media.netlify.app/image/202112/jet_plane_diagram.png)

Step 2: Derive equations

Using trigonometry, we can derive two equations:

tan(60) = height / distance1

tan(30) = height / distance2

where height is the constant height of the jet plane, and distance1 and distance2 are the distances travelled by the jet plane in the two different positions.

Step 3: Solve for unknowns

Using the two equations above, we can solve for the two unknowns:

distance1 = height / tan(60)

distance2 = height / tan(30)

Using the formula for average speed:

average speed = total distance / total time

We can calculate the total distance as:

total distance = distance1 + distance2

The total time is given as 15 seconds. Therefore, we can calculate the average speed of the jet plane as:

average speed = (distance1 + distance2) / 15

Step 4: Simplify

Substituting the values of distance1 and distance2, we get:

average speed = (height / tan(60) + height / tan(30)) / 15

Simplifying further, we get:

average speed = height(2 / √3 + 1) / 15

Substituting the value of height as 1500√3m, we get:

average speed = 500(2 + √3) m/s

Step 5: Final Answer

Therefore, the speed of the jet plane is 500(2 + √3) m/s.

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