The velocity of a 55.0-kg person hitting the ground, is mathematically given as
vt=39.5983m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
Generally, the equation for is mathematically given as
mass of squirrel,
[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]
Terminal velocity is given by:
[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]
where \rho is the density of fluid that is falling and it is given by
[tex]$\rho=\frac{m}{V}$[/tex]
since, volume =area * height
[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]
A is the surface area of squirrels.
C is the drag coefficient.
The surface area facing the fluid is given by:
[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]
so, terminal velocity is :
[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]
Vt=39.5983
In conclusion, the terminal velocity of the squirrel is 39.5983m/s
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