If a basketball is dropped from a helicopter, its velocity as a function of time v(t) can be modeled by the equation: v(t) = Squareroot 2 mg/rho AC_d (1 - e^-Squareroot rho g C_d A/2m t) where g = 9.81 m/s^2 is the gravitation of the Earth, C_d = 0.5 is the drag coefficient, rho = 1.2 kg/m^3 is the density of air, m is the mass of the basketball in kg, and A = pi r^2 is the projected area of the ball (r = 0.117 m is the radius). Note that initially the velocity increases rapidly, but then due to the resistance of the air, the velocity increases more gradually. Eventually the velocity approaches a limit that is called the terminal velocity. Determine the mass of the ball if at t = 5 s the velocity of the ball was measured to be 19.5 m/s.
