Question

81. A 560-g squirrel with a surface area of 930cm2 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?

Answer 1

Answer:

Explanation:

Given

mass of squirrel $m=560 gm$

Surface area of squirrel $A=930 cm^2$

and the area which face $A_f=\frac{A}{2}=465 cm^2$

height of tree $h=5 m$

Coefficient of drag $C=1$

drag Force $F_d=\frac{1}{2}C\cdot \rho \cdot A\cdot v^2$

Terminal velocity is given

$F_d=mg$

$\frac{1}{2}C\cdot \rho \cdot A\cdot v^2=mg$

$v=\sqrt{\frac{2\times m\times g}{\rho \times C\times A_f}}$

$v=\sqrt{\frac{2\times 0.560\times 9.8}{1.2\times 1\times 465\times 10^{-4}}}$

$v=13.9 m/s$

(b)Mass of person $m=56 kg$

$v^2-u^2=2gh$

$u=0$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 5}$

$v=9.89 m/s$

Answer 2

Answer:

(a). The terminal velocity is 9.83 m/s.

(b). The velocity is 9.9 m/s.

Explanation:

Given that,

Mass of squirrel = 560 g

Surface area = 930 cm²

Height = 5.0 m

Weight of person = 56 kg

We need to calculate the terminal velocity

Using formula of terminal velocity

$v_{t}=\sqrt{\dfrac{2mg}{\rho A C_{d}}}$

Where, A = area

$\rho$ = density

$C_{d}$ = drag coefficient

m = mass

Put the value into the formula

$v_{t}=\sqrt{\dfrac{2\times0.560\times9.8}{1.22\times0.093\times1}}$

$v_{t}=9.83\ m/s$

The terminal velocity is 9.83 m/s.

We need to calculate the velocity

Using equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v=\sqrt{2\times 9.8\times5}$

$v=9.9\ m/s$

Hence, (a). The terminal velocity is 9.83 m/s.

(b). The velocity is 9.9 m/s.