In February 1955, a paratrooper fell 370 mfrom 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 56m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the snow was at the survivable limit of$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{x}{\mathbf{10}}^{\mathbf{5}}$. 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?

The mass of Paratrooper with gear is,$m=85\mathrm{kg}$.

The speed at the impact is,$v_i=56\mathrm{m}/\mathrm{s}$.

The height of the plane is,$h=370\mathrm{m}$.

The magnitude of the force by the snow is,${\mathrm{F}}_{\mathrm{snow}}=1.2\times {10}^5\mathrm{N}$.

We can use the equation of impulse related to the change in momentum to calculate the impulse from the snow. The acceleration by the snow can be calculated using the kinematic equation which is given as.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ \mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ad} \end{gathered}$$

Let us apply Newton’s second law to the paratrooper when she hits the snow.

$\mathrm{mg}-{\mathrm{F}}_{\mathrm{snow}}=\mathrm{ma}$

Solving this equation for acceleration, we get

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{mg}-{\mathrm{F}}_{\mathrm{snow}}}{\mathrm{m}} \\ =\frac{85\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2-1.2\times {10}^5\mathrm{N}}{85\mathrm{kg}} \\ =-1402\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Now, we have, a kinematic equation as

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ad}$

Using this equation to calculate the depth of snow, we get

$$\begin{gathered} \mathrm{d}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2}{2\mathrm{a}} \\ =\frac{{\left(0\mathrm{m}/\mathrm{s}\right)}^2-{\left(56\mathrm{m}/\mathrm{s}\right)}^2}{-2\times 1402\mathrm{m}/{\mathrm{s}}^2} \\ =1.1\mathrm{m} \end{gathered}$$

Using the equation of impulse-momentum theorem, the impulse by the snow is,

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ =\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ =85\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-56\mathrm{m}/\mathrm{s}\right) \\ =-4.8\times {10}^3\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the magnitude of impulse is,

$J=4.8\times {10}^3\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$