Question

A 575 g squirrel with a frontal surface area of 0.0146 m^2 falls from a 8.5 m tree to the ground. Assume the density of air in this problem is given by 1.21 kg/m^3. (a) What is its terminal velocity in m/s? (Use a drag coefficient of 0.70 and assume down is positive) (b) What will the velocity (in m/s) of a 56 kg person falling that distance, assuming no drag contribution?

Answer 1

a) 30.20 m/s

b) 12.91 m/s

Step-by-step explanation:

Mass of squirrel = 575 g = m

Drag coefficient = 0.70 = C

Density of air = 1.21 kg/m³ = ρ

frontal surface area = 0.0146 m² = A

Height the squirrel falls = 8.5 m = h

a) Drag force

`F=(1)/(2)Av^2C\rho\\\Rightarrow F=(1)/(2)0.0146v^2* 0.7* 1.21`

This force will oppose gravity

`mg=F\\\Rightarrow 0.575* 9.81=(1)/(2)0.0146v^2* 0.7* 1.21^2\\\Rightarrow (0.575* 9.81* 2)/(0.0146* 0.7* 1.21)=v^2\\\Rightarrow v=30.20\ m/s`

∴ Terminal velocity is 30.20 m/s

b) Neglecting drag force we get

`mgh=(1)/(2)mv^2\\\Rightarrow v=√(2gh)\\\Rightarrow v=√(2* 9.81* 8.5)\\\Rightarrow v=12.91\ m/s`

∴ Velocity of a 56 kg person falling that distance, assuming no drag contribution is 12.91 m/s