- #1

Jazz

- 103

- 5

If you fall from a ##5\!-\!m## high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.

(Source: http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.9:33/College_Physics)

1. Homework Statement

1. Homework Statement

A squirrel (with ##0.0155\ m^2## of its body facing the fluid) falls from a ##5.0\!-\!m## tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a ##56\!-\!kg## person hitting the ground, assuming no drag contribution in such a short distance?

Given/Known Data:

##m_{squirrel} = 0.560\ kg##

##m_{person} = 56\ kg##

##A= 0.0155\ m^2##

##C_d= 1.0##

##\rho_{air} = 1.21\ kg/m^3##

##h = 5.0\ m##

## Homework Equations

##v_t = \sqrt{\frac{2mg}{C\rho A}}##

##v=\sqrt{2gh}##

## The Attempt at a Solution

After solving both cases from the equations above I get:

##v_t## of squirrel ##= 24.2\ m/s##

##v_t = \sqrt{\frac{2(0.560\ kg)(9.8\ m/s^2)}{(1.0)(1.21\ kg/m^3)(0.0155\ m^2)}}##

##v_{person} = 9.90\ m/s##

And in order to use the ##m_{person}## not needed in the previous questions and asumming an area of ##0.70\ m^2## their ##v_t## is ##36.0\ m/s##.

This seems correct, but my question is how does terminal velocity fit in all of this?

I mean that the justification that the squirrel gets undamaged after falling to the ground because it reaches its terminal velocity before than a human being doesn’t convince me. Clearly the squirrel doesn't hit the ground at ##24.2\ m/s## but how can I find its velocity at ##h=0##?

I guess this can be better explained by the change in momentum and/or the kinetic energy of the systems, but I cannot see a way of explaining it with the concepts I’ve learned so far (in this case, Newton’s Second Law and Drag Force (without Bouyant Force)).

Thanks!