Question

1.The angle of elevation of a jet plane from a point R on the ground is 60°. After a flight of

17 seconds, the angle of elevation changes
to 30°. If the jet plane is flying at a constant height of 1700√3 metres, find the speed of
the jet plane in km per hour.
(ans: 720km/hr)​

Answer 1

Diagram is not so perfect

OR = distance travelled by the jet plane.

Explanation:

Let point J be the jet plane and OJ be the constant height at jet plane is flying and OR be the distance jet plane is travelling.

h =

`1700 √(3)`

Now,

From the figure i drew (1st)

taking 60 as reference angle,

tan60° = p/b

or,

`tan60 = 1700 √(3) / b`

`b \: = 1700 √(3) / tan60`

so, b = 1700m

inital distance = 1.7km(OR)

now,

tan 30 = p/b

`tan30 = 1700 √(3) / b \\ or \: b = 1700 √(3) / tan30`

so, b = 5100 m

so, final distance = 5.1 km(OR)

Now,

distance travelled = final distance - initial distance

distance travelled = 5.1 km - 1.7km

distance travlled = 3.4 km

Now,

time = 17 seconds

= 0.00472 hour.

Now,

speed = d/t

or, speed = 3.4/0.00472

so, speed = 720.3389km/hr

so, speed = 720 km/hr approximately