Answer:
The time taken for the airplane to climb to a height of 12,500 feet is 2.83 minutes.
Explanation:
Given that,
Angle of the plane of ascent, [tex]\theta=16^{\circ}[/tex]
Initial speed of the plane, u = 267 ft/s
We need to find the time taken for the airplane to climb to a height of 12,500 feet. First lets find the vertical speed of the plane.
[tex]u_y=u\ \sin\theta\\\\u_y=267\times \ \sin(16)\\\\u_y=73.59\ ft/s[/tex]
Let t is the time taken for the airplane to climb to a height of 12,500 feet. The speed of an object is given by :
[tex]u=\dfrac{d}{t}\\\\t=\dfrac{d}{u}\\\\t=\dfrac{12500\ ft}{73.59\ ft/s}\\\\t=169.86\ seconds\\\\t=2.83\ min[/tex]
So, the time taken for the airplane to climb to a height of 12,500 feet is 2.83 minutes. Hence, this is the required solution.
