Question

(1 point) An object of mass 5 kg is given an upward initial velocity of 16 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -50, where v is the velocity of the object in m/sec. Assume gravitational constant is g = 9.8m/seca. Set up the differential equation for this scenario: v' = m/sec Solve the differential equation for the equation of motion: The equation is both separable and linear. Suppose we separate the variables, we will get dt = dv. We shall incorporate initial data using definite integrals, integrating from initial values (to = 0,v0 = 16) to current values (t, v): So 1 dx = Sie dy, The immediate result of the integrals is t= Solve for vi v(t) = m/sec.

Answer 1

tiren Mass of object = m = 5 kg Initial velocity of object = Vo = 16 m/s Magnitude of force of air resistance on object = Fr= 5V I acceleration due to gravity g = 9.8 m/s2 free body diagram of object can be drawn as follows Fy = mg where mga force of gravity (Fg) acting downward Force of resistance acting down wa a acceleration of object (upward) vo initial velocity of object Now, equation of motion for object is - o - Fg - Fr = ma =) -mg - 5 V = ma =) - 59 - 5V = sa ->) - 9-V = a -9-V = dy [: a = cant where vo velocity 7 3) V!= dy = -9-v --... 0 -> v' = -g-V ł i CH t time I 1 - - - - ñ ñ Ferom o er -9-V = at

limits to=o to t ② I Now, and integrating Vo=16 } in , we the get to v Id = av = - 1 dt dv 9 + V J+V 11 dx = x and dx = -en (a +) ...☺ By using ① , ② becomes (a=g and x =V) [+).* = [- en (9 +V)]," =) -o = - In 19 +v) - (- en (9 +16)) =) t = -en (9 + v) + en 19+16) = t = eng +1671 i en a- enb= ln (9)] LATV] - .-.- © Solving for V = V(+) By taking exponent on both sides of ③ we get et = [: eena = a 7 => (9 +V)et = g + 16 => get + vet = 9+16 = vet = 9+16 - get =) v = ó 69 +16 - get) V(+) = (9 +) + - 4 =) V(+) = -9 + (8 +16) et =) Vlt) = -9.8 +19.8+16)e-* =) V[+) = (-9.8 + 25.8et) mis (Final answer) ł