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 (6000m), 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 of3.0m.

Short answer

The deceleration was $-486\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Stating given data

• Initial velocity, U =$\mathbf{54}\mathrm{m}/\mathrm{s}$.
• Final velocity, V =0.
• Height, D, at which the trees and the snow stopped the pilot = $\mathbf{3}.\mathbf{0}\mathbf{m}$.

Calculation of deceleration of pilot

The velocity decreases; hence it is a case of deceleration.

When there is a change in the velocity, whether it is due to change in magnitude or direction, we can calculate the acceleration from the equation of motion as

$V^2-U^2=2ad$

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance that was covered to achieve the final velocity.

Substituting the values in the above expression, we get

$\begin{array}{c}(0{)}^2-(54{)}^2=2\times (3)\times a\\ -2916=2\times (3)\times a\\ a=\frac{-2916}{6}\\ a=-486\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence the deceleration is$-\mathbf{486}\mathrm{m}/{\mathrm{s}}^2$.