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?
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
distance flown is
d/11000 = csc 15°
so, time needed is d/265 seconds
Thanks so much!
To find the time it will take for the airplane to climb to a height of 11,000 feet, we need to use trigonometry and the given information.
First, we need to determine the vertical component of the plane's velocity. The vertical component can be found by multiplying the speed of the plane by the sine of the angle of ascent.
Vertical component of velocity = speed of the plane * sin(angle of ascent)
Vertical component of velocity = 265 ft/s * sin(15 degrees)
Now, we can calculate the time it will take for the airplane to climb to a height of 11,000 feet by dividing the height by the vertical component of velocity.
Time = height / vertical component of velocity
Time = 11,000 ft / (265 ft/s * sin(15 degrees))
To calculate this, we need to convert 11,000 feet to the same units as the vertical component of velocity. Let's convert it to seconds.
11,000 ft * 1 s/265 ft = 41.5094 seconds
Now, let's convert seconds to minutes.
41.5094 s * 1 min/60 s = 0.6918 min
Therefore, it will take approximately 0.6918 minutes (or about 41.5 seconds) for the airplane to climb to a height of 11,000 feet.