An airplane takes off at a 15° angle and travels at the rate of 350 ft/sec.

Question

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 35,000 feet? (Round your answer to two decimal places.)

min=

Answer 1

Given: Plane takes off at a 15 degree angle, at a speed of 350 feet/sec

Find: The time to reach an elevation of 35,000 feet.

Solution:

Visualize a right triangle in Q1 with 15 degree at the origin.

The hypotenuse is the path of the airplane.

Sin(15) = opposite / hypotenuse

Sin(15) = 35000 / (350*t)

350*t = 35000 / Sin(15)

t = 35000 / (350 * Sin(15) )

t = 100 / Sin(15)

t = 100 / .25882

t = 386.37 seconds

t = (386.37 sec) * (1 min. / 60 sec) = 6.44 minutes

Answer 2

Well vertical speed will be 350 sin 15° (ft/sec)

So divide 35000 by this, giving you the time in seconds