Question

In February 1955, a paratrooper fell 362 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 86 kg, and that the force on him from the snow was at the survivable limit of 1.2 ✕ 105 N.What is the minimum depth of snow that would have stopped him safely?

Answer 1

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m