Complete Question
A 560-g squirrel with a surface area of [tex]960 cm^2[/tex]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:
The terminal velocity of the 560-g squirrel is [tex]v = 9.9 \ m/s[/tex]
The velocity of the 56-kg person is [tex]v_p^2 = 9.9[/tex]
Explanation:
From the question we are told that
The mass of the squirrel is [tex]m_s =560 \ g = 0.56 \ kg[/tex]
The height of the fall is [tex]s = 5 \ m[/tex]
The drag coefficient of a skydiver is is [tex]C = 1[/tex]
The surface area is [tex]A = 930 \ cm^2 = 0.093 \ m^2[/tex]
Generally the density of air is [tex]\rho = 1.21 \ kg /m^3[/tex]
Generally the terminal velocity is mathematically represented as
[tex]v = \sqrt{ \frac{2 * m * g}{ \rho * C *A } }[/tex]
=> [tex]v = \sqrt{ \frac{2 * 0.56 * 9.8}{ 1.21 * 1 * 0.093 } }[/tex]
=> [tex]v = 9.9 \ m/s[/tex]
Generally from kinematic equation
[tex]v_p^2 = u^2 + 2gs[/tex]
given that the person was at rest before the fall u = 0 m/s
[tex]v_p^2 = 0+ 2* 9.8 * 5[/tex]
=> [tex]v_p^2 = 9.9[/tex]
