In february 1955, a paratrooper fell 365 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 56 m/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 1.2 x 10^5 n.what is the minimum depth of snow that would have stopped him safely? what is the magnitude of the impulse on him from the snow?

a.

b.

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

b)

The momentum is equal to the impulse on him from the snow so:

    vₐ = v_c  

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c =

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( )

            R ’= R + ΔR = R ( )

we substitute

           vₐ =

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( )

we make the product and keep the terms linear

        vₐ = v_c