An object of mass 20 kg is released from rest 4000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant bequals50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground? Assume that the acceleration due to gravity is 9.81 m divided by sec squared and let x(t) represent the distance the object has fallen in t seconds.
Accepted Solution
A:
Answer:t = 1019.76 secStep-by-step explanation:object mass = 20 kgreleased from rest = 4000 m aboveproportionality constant(b) = 50 N-sec/minitial velocity = v₀we know,[tex]v(t) = \dfrac{mg}{b}+(v_0- \dfrac{mg}{b})e^{\dfrac{-bt}{m}}[/tex]and[tex]x(t) = \dfrac{mg}{b}t+\dfrac{m}{b}(v_0- \dfrac{mg}{b})(1-e^{\dfrac{-bt}{m}})[/tex][tex]x(t) = \dfrac{20\times 9.81}{50}t+\dfrac{20}{50}(0- \dfrac{20\times 9.81}{50})(1-e^{\dfrac{-50t}{20}})[/tex][tex]x(t) = 3.924 t - 1.5696(1-e^{-2.50t})[/tex]x(t) = 4000[tex]4000 = 3.924 t - 1.5696(1-e^{-2.50t})[/tex]neglecting [tex]e^{-2.50t}[/tex] we get4000 = 3.924 t - 1.5696t = 1019.76 sec