When a base jumper leaps from a tall cliff, the distance she falls (in meters) after t seconds, accounting for air resistance, is gien by d(t)equals=startfraction m over k endfraction ln [hyperbolic cosine (startroot startfraction kg over m endfraction endroot t )]m klncosh kg mt, where m is her mass in kilograms, gequals=9.8 m divided by s squared9.8 m/s2 is the acceleration due to gravity, and k is a physical constant. her velocity t secondst seconds after jumping is v(t)equals=d prime (t )d′(t)equals=startroot startfraction mg over k endfraction endroot hyperbolic tangent (startroot startfraction kg over m endfraction endroot t )mg ktanh kg mt. complete parts (a) through (d) below. a. compute the jumper's terminal velocity, which is defined as modifyingbelow lim with t right arrow infinitylimt→[infinity]v(t).