In December of 1989, a KLM Boeing 747 airplane carrying 231 passengers entered a cloud of ejecta from an Alaskan volcanic eruption. All four engines went out, and the plane fell from 27800 ft to 13700 ft before the engines could be restarted. It then landed safely in Anchorage. Neglecting any air resistance and aerodynamic lift, and assuming that the plane had no vertical motion when it lost power, (a) for how long did it fall before the engines were restarted, and (b) how fast was it falling at that instant?
a. d = Vo*t + 0.5g*t^2 = 27800-13700,
0 + 16.2*t^2 = 14100,
t^2 = 872,
t = 29.5 s.
b. Vf^2 = Vo^2 + 2g*d.
Vf^2 = 0 + 64.7*14100 = 912270,
Vf = 955 Ft/s.
To solve this problem, we can use the equations of motion to find the time and speed at the instant the engines were restarted.
(a) First, let's find the time it took for the plane to fall before the engines were restarted. We can use the equation for distance fallen without air resistance:
h = 0.5 * g * t^2
Where:
h = height fallen (initial altitude - final altitude) = 27800 ft - 13700 ft = 14100 ft
g = acceleration due to gravity = 32.2 ft/s^2 (neglecting any variations)
Rearranging the equation to solve for time:
t^2 = (2 * h) / g
t^2 = (2 * 14100 ft) / 32.2 ft/s^2
t^2 = 876.95 s^2
Taking square root on both sides:
t = 29.62 seconds (approximately)
So, it took approximately 29.62 seconds for the plane to fall before the engines were restarted.
(b) Now, let's find the speed of the plane at the instant the engines were restarted. We can use the equation for velocity:
v = g * t
Substituting the values:
v = 32.2 ft/s^2 * 29.62 s
v = 953.28 ft/s (approximately)
Therefore, the speed of the plane at the instant the engines were restarted was approximately 953.28 ft/s.