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I'm not really sure how to approch this question. I think simply it would look like a right angles triangle, I would need an initial velocity (ie just at take off) to work it out?

Thanks for your help!

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- Thread starter Rebecca_M
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Solving for x, we get x = 10.7/tan(2.23) = 288.3 meters.In summary, the question was how to find the horizontal distance traveled by an aircraft just after take off to a vertical height of 10.7m, with a rate of climb of 1920m/min. The approach involved finding the angle of takeoff and using trigonometric functions to calculate the horizontal distance. However, since the angle of takeoff was not given, the question could not be solved.

- #1

- 2

- 0

I'm not really sure how to approch this question. I think simply it would look like a right angles triangle, I would need an initial velocity (ie just at take off) to work it out?

Thanks for your help!

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- #2

Homework Helper

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It is very important to post the entire question, word for word.

In this case, the angle of takeoff is missing so it cannot be solved.

If, for example, the plane was a Harrier and took off vertically, the horizontal distance would be zero. If it went up at 45 degrees, the horizontal distance would be 10.7 meters.

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- #4

Homework Helper

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The total speed is 385 km/h = 385*1000m/3600s = 107 m/s.

The angle above horizontal must be θ where sin(θ) = 4.17/107 so θ=2.23 degrees. Using the vertical distance of 10.7 m, we have

tan(2.23) = 10.7/x where x is the horizontal distance.

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