Question

In February 1955, a paratrooper fell 1200 ft from an

airplanewithout being able to open his chute but happened to land
in snow,suffering only minor injuries. Assume that his speed at
impact was56 m/s (terminal speed0, that his mass (including gear)
was 85 kg,and that the force on him from the snow was at the
survivable limitof 1.2 X 10 ^ 5 N. What is the minimum depth of
snow that wouldhave stopped him safely?

Answer 1

`s=1.1107\ m` is the minimum depth of snow for survivable stopping.

Step-by-step explanation:

Given:

• terminal velocity of fall,
  `u=56\ m.s^(-1)`

• mass of the paratrooper,
  `m=85\ kg`

• force on the paratrooper by the ice to stop him,
  `F=1.2* 10^5\ N`

Firstly, we calculate the deceleration caused in the snow:

`a=(F)/(m)`

`a=(120000)/(85)`

`a=1411.765\ m.s^(-2)`

Now, using equation of motion:

`v^2=u^2+2a.s` ....................(1)

where:

v = final velocity of the body after stopping

u = initial velocity of the body just before hitting the snow

a = acceleration of the body in the snow

s = distance through in the snow

Putting respective values in eq. (1)

`0^2=56^2+2* (-1411.765)* s`

`s=1.1107\ m`