Question

In February 1955, a paratrooper fell 370 m from an airplane withoutbeing able to open his chute but happened to land in snow,suffering only minor injuries. Assume that his speed at impact was45 m/s (terminal speed), that his mass (including gear) was 65 kg,and that the magnitude of the force on him from the snow was at thesurvivable limit of 1.4 x 105 N. What are (a) the minimum depth ofsnow that would have stopped him safely and (b) the magnitude ofthe impulse on him from the snow?

Answer 1

survivable limit of 1.4 x 105 N

F= 1.4 x 10⁵ N

retardation =a= F/mass of body

               = 1.4 x 10⁵ /65

                =2153.84 m/s²

a) depth of snow , = y

    using newtons 3rd law we get

    v² -u² =2 * a *s

where s= y

a= retadation which is evaluate above

u= 0 ,i.e initial velocity

v=45m/s

so we get y,

y= v² /(2 * a)

=0.47 m in snow

b)

impulse = change in momentum

= m( 45-0)

= 65 * 45 =2925 kg m/s