Answer:
a) 30.20 m/s
b) 12.91 m/s
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
[tex]F=\frac{1}{2}Av^2C\rho\\\Rightarrow F=\frac{1}{2}0.0146v^2\times 0.7\times 1.21[/tex]
This force will oppose gravity
[tex]mg=F\\\Rightarrow 0.575\times 9.81=\frac{1}{2}0.0146v^2\times 0.7\times 1.21^2\\\Rightarrow \frac{0.575\times 9.81\times 2}{0.0146\times 0.7\times 1.21}=v^2\\\Rightarrow v=30.20\ m/s[/tex]
∴ Terminal velocity is 30.20 m/s
b) Neglecting drag force we get
[tex]mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 8.5}\\\Rightarrow v=12.91\ m/s[/tex]
∴ Velocity of a 56 kg person falling that distance, assuming no drag contribution is 12.91 m/s
