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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?

Respuesta :

davidlcastiblanco

Answer:

a. [tex]i=4760 kg*m/s[/tex]

b. [tex]D_U= 1.11 m[/tex]

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

[tex]K_E= (1/2)*85kg*(56m/s)^2[/tex]

[tex]K_E=133280 J[/tex]

[tex]D_U= \frac{133280J}{120000N} = 1.11 m[/tex]

b)

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

[tex]p=m*v[/tex]

[tex]p=85kg*56m/s[/tex]

[tex]i=4760 kg*m/s[/tex]

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