In February 1955, a paratrooper fell 370 m from an airplane without being able t
ID: 1570807 • Letter: I
Question
In February 1955, a paratrooper fell 370 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 59 m/s (terminal speed), that his mass (including gear) was 68 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.5 times 10^5 N. What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow? (a) Number Units (b) Number UnitsExplanation / 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 1.5e5 N / 68 kg, or 2.2e3 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)
= .79 m
The impulse would simply be the change in momentum, or
68 kg * 59 m/s = 4.012e3 kg-m/s. This is a lot easier than calculating the time it would take to stop the paratrooper and then multiply the force by that number.