Asked by Arora
When an airplane leaves the runway, its angle of climb is 24° and its speed is 345 feet per second. How long will it take the plane to climb to an altitude of 10,000 feet? 16,000 feet? (Round your answers to one decimal place.)
10,000 ft = ? in s
16,000 ft = ? in s
10,000 ft = ? in s
16,000 ft = ? in s
Answers
Answered by
Henry
Vo = 345Ft/s[24o].
Xo = 345*Cos24 = 315 Ft/s.
Yo = 345*sin24 = 140 Ft/s.
Y^2 = Yo^2 + 2g*h.
0 = 140^2 - 64*h. h = 306.25 Ft., max.
Based on the information given, the plane can reach a maximum ht. of only 306 Ft.
Xo = 345*Cos24 = 315 Ft/s.
Yo = 345*sin24 = 140 Ft/s.
Y^2 = Yo^2 + 2g*h.
0 = 140^2 - 64*h. h = 306.25 Ft., max.
Based on the information given, the plane can reach a maximum ht. of only 306 Ft.
Answered by
Henry
Correction: The plane is not a free-fall object.
a. d = Yo*t.
10,000 = 140*t, t = ?.
b. 16,000 = 140*t, t = ?.
a. d = Yo*t.
10,000 = 140*t, t = ?.
b. 16,000 = 140*t, t = ?.
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