Parachutists whose chutes have failed to open have been known to survive if they land in deep snow. Assume that a 75-kg parachutist hits the ground with an area of impact of\(0.30\;{{\rm{m}}^2}\)at a velocity of\(55\;{\rm{m/s}}\)and that the ultimate strength of body tissue is\(5 \times {10^5}\;{\rm{N/}}{{\rm{m}}^2}\). Assume that the person is brought to rest in 1.0 m of snow. Show that the person may escape serious injury.

The initial velocity of the parachutist is\(u = 55\;{\rm{m/s}}\).

The body tissue strength is\({\sigma _{\rm{t}}} = 5 \times {10^5}\;{\rm{N/}}{{\rm{m}}^2}\).

The area of impact is\(A = 0.30\;{{\rm{m}}^2}\).

The mass of the parachutist is\(m = 75\;{\rm{kg}}\).

The height to which the parachutist seeps in the snow is\(s = 1\;{\rm{m}}\).

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it.The free body shown below depicts the forces on the parachutist.

The forces acting on the parachutist are the following:

• Weight of the parachutist,\(W = mg\)vertically downwards
• Force of the snow, \({F_{{\rm{snow}}}}\) on the parachutist

Using the kinematic equation of motion gives the average acceleration of the parachutist.

\(\begin{array}{c}{v^2} - {u^2} = 2as\\0 - {\left( {55\;{\rm{m/s}}} \right)^2} = 2a\left( { - 1\;{\rm{m}}} \right)\\a = \frac{{{{\left( {55\;{\rm{m/s}}} \right)}^2}}}{{2\left( {1\;{\rm{m}}} \right)}}\\ = 1513\;{\rm{m/}}{{\rm{s}}^2}\end{array}\)

According to Newton’s second law of motion,

\(\begin{array}{c}{F_{{\rm{snow}}}} - mg = ma\\{F_{{\rm{snow}}}} = m\left( {a + g} \right)\\ = \left( {75\;{\rm{kg}}} \right)\left( {1513\;{\rm{m/}}{{\rm{s}}^2} + 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ = 1.14 \times {10^5}\;{\rm{N}}\end{array}\)

Stress is defined as the force experienced by a body per unit area. The formula for stress is

\(\sigma = \frac{{{\rm{Force}}}}{{{\rm{Area}}}}\).

In this case, the\({F_{{\rm{snow}}}}\)produces stress on the parachutist.

\(\begin{array}{c}\sigma = \frac{{1.14 \times {{10}^5}\;{\rm{N}}}}{{{\rm{0}}{\rm{.30}}\;{{\rm{m}}^2}}}\\ = 3.8 \times {10^5}\;{\rm{N/}}{{\rm{m}}^2}\end{array}\)

The calculated stress is less than the ultimate tissue strength of the body, i.e.,

\(\sigma < {\sigma _{\rm{t}}} = 5 \times {10^5}\;{\rm{N/}}{{\rm{m}}^2}\).

Therefore, the parachutist can avoid serious injury. If he lands on his feet first, the legs may experience more than the average force and sustain injury.