Asked by Himanshu

The angle of elevation from Jet fighter. A ground on 60 degree after a flight takeoff 15 seconds the angle of elevation 30 degree if you at speed of 720 km per hour then find the constant height reached at the flying

Answers

Answered by Reiny
I will assume the following:
when t = 0 , angle of elevation is 60°
when t = 15 seconds, angle of elevation is 30°
horizontal speed is 720 km/h

let the constant height that the plane is flying be h m

I have a right-angled triangle with height h m, base of x m
and base angle of ذ degrees

speed of 720 km/h = 720000/3600 m/s = 200 m/s

tanØ = h/x -------> h = x tanØ

case 1: when Ø = 60° , tan 60° = √3 , height = h, base = x
h = x√3

case 2: when Ø = 30° , horizontal distance = x + 15(200) = x + 3000
tan30° = 1/√3
h = (x + 3000)(1/√3) = (x+3000)/√3

√(x+3000)/√3 = x√3
x+3000 = 3x
x = 1500

then in : h = x√3
h = 1500√3 or appr 2598 m or 2.598 km
Answered by Reiny
typo in 5th last line:
should be
<b>(x+3000)/√3 = x√3</b> , not √(x+3000)/√3 = x√3

has no effect on what follows
Answered by Steve
750 km/hr = 200 m/s
so, in 15 seconds, the plane has gone 3000m
Draw a diagram. It should be clear that if the plane's height is h, then
h cot30° - h cot60° = 3000
h = 3000/(√3 - 1/√3) = 1500√3 meters
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