A 575-g squirrel with a surface area of 915 cm 2 falls from a 5.2-m tree to the

Question

A 575-g squirrel with a surface area of 915 cm2 falls from a 5.2-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.4 cm and length 22.8 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
  m/s

What will be the velocity of a 57.5-kg person hitting the ground, assuming no drag contribution in such a short distance?
  m/s

Explanation / Answer

Given

Mass of the squirrel m = 575 g = 0.575 Kg

Surface area S = 915 cm2 = 915 x 10-4 m2

Height of the tree h = 5.2 m

Mass of the man M = 57.5 kg

Known

Drag coefficient of a Skydiver (horizontal) C = 1.0

Density of air = 1.225 kg/m3

Acceleration due to gravity g = 9.8 m/s2

Solution

1)

The cross sectional area of the squirrel = area of the rectangle of dimensions ( 11.4 cm x 22.8 cm)

A = 11.4 x 22.8

A = 259.92 cm2

A = 259.92 x 10-4 m2

Terminal velocity

v = [2mg/CA]1/2

v = [ 2 x 0.575 x 9.8 / 1.225 x 1.0 x 259.92 x 10-4]1/2

v = 353.961/2

v = 353.96

v = 18.81 m/s

2)

the potential energy of the man at height h is

Epotential = Mgh

This energy will be convered into kinetic enrgy when he falls

Ekinetic = ½ Mv2

as per the conservation of energy

Epotential = Ekinetic

Mgh = ½ Mv2

v = (2gh)

v = ( 2 x 9.8 x 5.2)

v = 10.1 m/s

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