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Ugo [173]
3 years ago
8

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 dra

g 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
Physics
1 answer:
bija089 [108]3 years ago
7 0

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

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