A jet climbs at an angle of 26 degrees to the ground. The cruising height for the jet is 8000m, the climbing speed is 300km/hr and the cruising speed is 450km/hr. How far is it from the airport 5 minutes after taking off? Please explain thoroughly.

it takes 8/300 hours to get to cruising height. That's 1.6 minutes

in the remaining 3.4 minutes it travels 3.4/60*450 = 25.5 km

during the climb, it traveled 8cot26° = 16.4 km along the ground.

After flying the additional 25.5 km, it is 41.9 km away along the ground.

So, the actual straight-line distance d is given by

d^2 = 41.9^2 + 8^2

To find the distance from the airport 5 minutes after taking off, we need to determine the vertical and horizontal components of the distance traveled by the jet during that time.

First, let's calculate the vertical component of the jet's climb. We know that the cruising height is 8000m, and the jet has been climbing at an angle of 26 degrees to the ground. We can use the concept of trigonometry to find the vertical component.

The vertical component (V) can be calculated using the formula: V = height * sin(angle), where height is the cruising height and angle is the angle of climb.

V = 8000 * sin(26°)
V ≈ 8000 * 0.4384
V ≈ 3507.2 meters

So, the vertical component of the distance traveled by the jet is approximately 3507.2 meters.

Next, let's calculate the horizontal component of the jet's climb. We know that the climbing speed is 300 km/hr, and to find the horizontal component, we need to convert this speed to meters per minute since we are dealing with a time of 5 minutes.

The horizontal component (H) can be calculated using the formula: H = speed * time, where speed is the climbing speed and time is the duration of climb.

Climbing speed in meters per minute = (300 km/hr * 1000 m/km) / 60 min
Climbing speed in meters per minute ≈ 5000 meters per minute

H = 5000 meters per minute * 5 minutes
H = 25000 meters

So, the horizontal component of the distance traveled by the jet during the climb is 25000 meters.

Now, let's add the vertical and horizontal components to get the total distance from the airport 5 minutes after taking off.

Total distance = sqrt(Vertical component^2 + Horizontal component^2)
Total distance = sqrt(3507.2^2 + 25000^2)
Total distance ≈ sqrt(12298576.8 + 625000000)
Total distance ≈ sqrt(637298576.8)
Total distance ≈ 25258.55 meters

Therefore, approximately 5 minutes after taking off, the jet is approximately 25258.55 meters (or 25.26 km) away from the airport.