91Ó°ÊÓ

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,${\mathrm{F}}_{\mathrm{snow}}=1.2×{10}^5\mathrm{N}$.

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.

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

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

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

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

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

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

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

So, the magnitude of impulse is,

$J=4.8×{10}^3 \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}$