Question

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectangular prism with cross-sectional area of width 11.6 cm and length 23.2 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.)

Answer 1

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=
`(530)/(1000)=0.530 kg`

1kg=1000 g

Area=A=
`935 cm^2=935* 10^(-4) m^2`

`1 cm^2=10^(-4) m^2`

Height,h=4.4 m

C=1

`\rho=1.21 kg/m^3`

Width of rectangular prism,b=11.6 cm=
`(11.6)/(100)=0.116 m`

1 m=100 cm

Length,l=23.2 cm=
`0.232 m`

Area=
`l* b=0.116* 0.232=0.0269 m^2`

Terminal velocity,
`v_t=\sqrt{(2mg)/(\rho CA)}`

Where
`g=9.8 m/s^2`

Using the formula

`v_t=\sqrt{(2* 0.530* 9.8)/(1.21* 1* 0.0269)}`

`v_t=17.86 m/s`

The velocity of person,
`v=√(2gh)`

Using the formula

`v=√(2* 9.8* 4.4)`

`v=9.29 m/s`