Question

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. 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.

(a) 2.8 m/s²

(b) 5.6 m/s²

(c) 9.8 m/s²

(d) 11.2 m/s²

Answer 1

The deceleration experienced by the pilot upon impact would be (c) 9.8 m/s².

Step-by-step explanation:

When the pilot jumps from the flaming airplane, his initial speed is given as 54 m/s

The final speed is brought to a stop by the trees and snow over a distance of 3.0 m.

To find the deceleration, we can use the kinematic equation:

[v_f^2 = v_i^2 + 2a d]

Where:

(v_f) = final speed (0 m/s, as the pilot comes to a stop),

(v_i) = initial speed (54 m/s),

(a) = acceleration (deceleration in this case),

(d) = distance (3.0 m).

Rearranging the equation to solve for \(a\):

\[a = \frac{v_f^2 - v_i^2}{2d}\]

Plugging in the values:

[a = frac{0 - (54)^2}{2 times 3.0}]

[a = -frac{2916}{6}

[a = -486 , text{m/s}^2]

The negative sign indicates deceleration.

However, since the question asks for the magnitude, we take the absolute value, which is 486 m/s².

This is approximately equal to 9.8 m/s².

In summary, the correct answer is (c) 9.8 m/s².