SOLUTION: An airplane takes off at a 15° angle and travels at the rate of 350 ft/sec. Approximately how long does it take the airplane to reach an altitude of 25,000 feet? (Round your answer
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Question 924982: An airplane takes off at a 15° angle and travels at the rate of 350 ft/sec. Approximately how long does it take the airplane to reach an altitude of 25,000 feet? (Round your answer to two decimal places.)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
If you draw out the triangle, you get
* y is the height of the airplane at time t (y is the opposite side)
* 350t represents how far along the hypotenuse the plane has traveled
* t is time in seconds
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From the drawing above, we see
opposite = y
hypotenuse = 350t
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So that means we can say
sin(theta) = opposite/hypotenuse
sin(15) = y/350t
sin(15) = y/350t
350t*sin(15) = y
(350*sin(15))*t = y
y = (350*sin(15))*t
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The function represents how high the plane is at time t.
y is the vertical distance.
t is time in seconds.
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Plug in y = 25,000 and solve for t
... use a calculator here (type in "25000/(350*sin(15))" without quotes).
Special Note: make sure you're in degree mode and NOT in radian mode.
... round to 2 decimal places
It takes approximately 275.98 seconds for the airplane to climb to 25,000 ft. So that is your answer.
Note: 275.98 seconds = 4 minutes, 35.98 seconds = 4.6 minutes (approximate). This is extra info to possibly think of the timeframe in an easier context.
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