In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot's speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Question

Jessie Hanson

Physics

30 January, 05:54

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot's speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

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Nolan Medina · Answer 1

The deceleration of the pilot was 1.9 x 10⁴ m/s²

Explanation:

First, let's calculate the velocity of the pilot 3 m above the ground. To do that, we need to know how much time the pilot was falling. The equation for the position and velocity of the pilot are as follows:

y = y0 + v0 * t + 1/2 * g * t²

v = v0 + g * t

where

y = position of the pilot at time t

y0 = initial position

v0 = initial velocity

t = time

v = velocity at time t

g = acceleration due to gravity

If we consider the ground as the center of our reference system and that the pilot fell with an initial velocity of 0 (the pilot would unlikely impulse himself to the ground), then the equations would be as follows:

y = y0 + 1/2 g * t²

v = g * t

The time at which the pilot was 3.0 m above the ground will be:

3.0 m = 6000 m - 1/2 * 9.8 m/s² * t²

3.0 m - 6000 m / - 4.9 m/s² = t²

t = 35.0 s

The velocity at that time will be:

v = - 9.8 m/s² * 35.0 s = - 343 m/s

After 35.0 s the pilot has a positive acceleration besides the acceleration due to gravity. Then, the equation for velocity and position will be:

v = v0 + a*t now, v0 = - 343 m/s and a ≠ g and a>0

y = y0 + v0 * t + 1/2 * a * t² now, y0 = 3 m

Again, let's find the time at which the pilot hits the ground:

v = v0 + a*t

v-v0 / t = a

Replacing in the equation for position:

y = y0 + v0 * t + 1/2 * ((v-v0) / t) * t²

y = y0 + v0 * t + 1/2 * v * t - 1/2 * v0 * t)

y = y0 + 1/2 v0 * t + 1/2 v * t

replacing with numbers:

0m = 3m + 1/2 * (-343 m/s) t + 1/2 * (-54 m/s) t

-3.0 m = - 198.5 m/s * t

t = - 3m / - 198.5 m/s

t = 0.015 s

the upward acceleration was then:

v-v0 / t = a

-54 m/s - (-343 m/s) / 0.015 s = a

a = 1.9 x 10⁴ m/s²