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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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- #1

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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!

- #2

Delphi51

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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

Delphi51

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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