# Find initial velocity using rate of climb

• Rebecca_M
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.

#### Rebecca_M

Hi I need to find the horizontal distance traveled by an aircraft just after take off to a vertical height of 10.7m. The rate of climb is 1920m/min.

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?

Welcome to PF, Rebecca.
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.

Thanks for your reply Delphi51, unfortunatly I didn't have a entire question, I was trying to preempt an exam question that I had the next day. The exam question ended up having at take off speed of 385Km/h (at point of lift off) and a rate of climb of 250m/min. I still don't really have an idea how it works.

Well, that is not perfectly clear but likely means the vertical component of speed is 250 m/minute = 4.17 m/s.
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.