a paratrooper fell 360 m from an airplane without being able to open his chute b

Question

a paratrooper fell 360 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 52 m/s (terminal speed), that his mass (including gear) was 80 kg, and that the force on him from the snow was at the survivable limit of 1.2 105 N.
(a) What is the minimum depth of snow that would have stopped him safely?
(b) What is the magnitude of the impulse on him from the snow?

Explanation / Answer

Assuming a constant deceleration, the depth to which the trooper penetrated the snow must be vf² = vo² + 2ay where vf is the final (zero) velocity, vo is the initial velocity, and y was the depth of penetration. The acceleration a will be F/m, where F = total force. This total acceleration will be 1.2e5 N / 80 kg, or 1.56e3 m/s². This also includes gravity, but the 10 m/s² or so that we must subtract for gravitional accereration is not enough to make a significant difference in the result and may be neglected. Solving for the minimum depth of penetration gives y > vo² / (2a) = 0.867 m The impulse would simply be the change in momentum, or 80 kg * 52 m/s = 4.16e3 kg-m/s

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