Explanation:
Terminal velocity is given by:
[tex]v_t=\sqrt{\frac{2mg}{\rho C_dA}}[/tex]
Here, m is the mass of the falling object, g is the gravitational acceleration, [tex]C_d[/tex] is the drag coefficient, [tex]\rho[/tex] 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:
[tex]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[/tex]
Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is [tex]1.28\frac{kg}{m^3}[/tex].
[tex]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}[/tex]
Without drag contribution the motion of the person is an uniformly accelerated motion, thus:
[tex]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}[/tex]
