Question
Question: The angle of elevation of a jet plane from a point A on the ground is \[{{60}^{\circ }}\]. After a f...
The angle of elevation of a jet plane from a point A on the ground is 60∘. After a flight of 15 seconds, the angle of elevation changes to 30∘. If the jet plane is flying at a constant height of 15003 meters, find the speed of the jet plane. (3=1.732)
(a)200 m/s
(b)600 m/s
(c)300 m/s
(d)400 m/s
Solution
Hint: Draw figure as mentioned. Consider the triangles formed, find the distance travelled by the jet plane in 15seconds. Thus apply the distance and time in the formula of speed and get speed to the jet plane.
Complete step-by-step answer:
We have been given the angle of elevation of a jet plane from point A on the ground.
The angle of elevation from point A to the jet plane of position D is 60∘. (From the figure)
Now the jet plane travels from point D to point E in 15 sections. Now as the jet plane is at E, the angle of elevation from ground point A to point E becomes 30∘.
The jet plane is flying at a constant height from the ground, which is 15003 meters.
From the figure we can say that DB = EC = 15003, i.e. at the height is constant from the ground and they are perpendicular to the ground.
Now, let us consider the ΔADB,
tan60∘ = opposite side / adjacent side = ABDB.
From the trigonometric table, we know that, tan60∘=3.
DB = 15003 (from figure)