Question

In February 1955, a paratrooper fell 1200 ft from an

Answer 1

Answer:

$s=1.1107\ m$ is the minimum depth of snow for survivable stopping.

Explanation:

Given:

• terminal velocity of fall, $u=56\ m.s^{-1}$

• mass of the paratrooper, $m=85\ kg$

• force on the paratrooper by the ice to stop him, $F=1.2\times 10^5\ N$

Firstly, we calculate the deceleration caused in the snow:

$a=\frac{F}{m}$

$a=\frac{120000}{85}$

$a=1411.765\ m.s^{-2}$

Now, using equation of motion:

$v^2=u^2+2a.s$ ....................(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)

$0^2=56^2+2\times (-1411.765)\times s$

$s=1.1107\ m$