Asked by Trig
After leaving a runway a planes angle of ascent is 15 degrees and its speed is 265 feet per second. How many minutes will it take for the airplane to climb to a height of 11,000 feet?
Answers
Answered by
Reiny
let the time taken be t minutes
velocity = 265 ft/sec = 15900 ft/min
distance traveled along the hypotenuse
= 15900t ft
Did you make a sketch of the right-angled triangle?
I see it as ...
sin 15° = 11000/15900t
15900t = 11000/sin15
t = 110/(159sin15) = 2.69 minutes
velocity = 265 ft/sec = 15900 ft/min
distance traveled along the hypotenuse
= 15900t ft
Did you make a sketch of the right-angled triangle?
I see it as ...
sin 15° = 11000/15900t
15900t = 11000/sin15
t = 110/(159sin15) = 2.69 minutes
Answered by
Steve
distance flown is
d/11000 = csc 15°
so, time needed is d/265 seconds
d/11000 = csc 15°
so, time needed is d/265 seconds
Answered by
Trig
Thanks so much!
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