Question

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 55 m/s (terminal speed), that his mass (including gear) was 68 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.1 × 105 N. What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

Answer 1

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$