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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?

Respuesta :

Answer:

[tex]s=1.1107\ m[/tex] is the minimum depth of snow for survivable stopping.

Explanation:

Given:

  • terminal velocity of fall, [tex]u=56\ m.s^{-1}[/tex]
  • mass of the paratrooper, [tex]m=85\ kg[/tex]
  • force on the paratrooper by the ice to stop him, [tex]F=1.2\times 10^5\ N[/tex]

Firstly, we calculate the deceleration caused in the snow:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{120000}{85}[/tex]

[tex]a=1411.765\ m.s^{-2}[/tex]

Now, using equation of motion:

[tex]v^2=u^2+2a.s[/tex] ....................(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)

[tex]0^2=56^2+2\times (-1411.765)\times s[/tex]

[tex]s=1.1107\ m[/tex]

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