Question

You are stationed at a radar base and you observe an unidentified plane at an altitude h = 5000 m
flying towards your radar base at an angle of elevation = 30°. After exactly one minute. your radar sweep
reveals that the plane is now at an angle of elevation = 60° maintaining the same altitude. What is the
speed (in m s) of the plane?
​

Answer 1

Answer:

5773.502691896257

Step-by-step explanation:

tan 30 = DC/ AC = 5000/AC

1/√3= 5000/AC

AC = 5000√3

tan 60 = ED/ AB = 5000/AB

√3 = 5000/AB

AB = 5000/ √3

DE = AC- AB = 5000√3 - 5000 / √3

= 10000√3 m

Speed = distance/ time

= 10000√3 m/ 60 secs

= 5773.502691896257

Answer 2

Answer:

Speed = 1000 m/s approx

Step-by-step explanation:

when the angle between the radar base and the plan is 30°and the height is 5000m

let the angle ∠ABC be 30° and AC be 5000m

there fore tan 30° = $\frac{AC}{BC}$

                             $\frac{1}{\sqrt{3} }$  = $\frac{5000}{x}$

                              x= 5000$\sqrt{3}$ m

When the angle between the radar base and the plan is 60°and the height is 5000m

let the angle ∠A'B'C' be 30° and A'C' be 5000m

there fore tan 60° = $\frac{A'C'}{B'C'}$

                           $\sqrt{3}$ = $\frac{5000}{x'}$

                            x' = $\frac{5000}{\sqrt{3} }$

Distance is  =  5000$\sqrt{3}$ m-  $\frac{5000}{\sqrt{3} }$

                    = $\frac{10000}{\sqrt{3} }$m

Speed = $\frac{Distance}{Time}$

           = $\frac{10000}{\sqrt{3}*60}$

          = $\frac{10000}{10.38}$

          = 1000 m/s approx