Question

An object of mass 450 kg is released from rest 2000 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 bequals 50 ​N-sec/m, determine the equation of motion of the object. When will the object strike the​ ground? [Hint: Here the exponential term is too large to ignore. Use​ Newton's method to approximate the time t when the object strikes the​ ground.] Assume that the acceleration due to gravity is 9.81 m divided by sec squared .

Answer 1

Answer:

The answer to the question is;

The equation of motion of the object is

x(t) = 88.29·t + 794.6$e^{-\frac{t}{9} }$  -794.6

Explanation:

The force due to air resistance = 50 N-sec/m

dividing by the mass we get 50 N-sec/m /(450 kg) = 1/9 m-s/s²m =1/9 s⁻¹

Therefore   F₁ = 450*9.81 = 4414.5 N, F₂ = -bv(t) = -50v

Formula is $m\frac{dv}{dt} = mg-bv(t)$

$450\frac{dv}{dt} = 450*9.81-50v$

$\frac{dv}{dt} = 9.81-v/9$

$\int\limits^.\frac{dv}{9.81-v/9} = \int\limits^ .dt$

=$-9\int\limits^.\frac{du}{u} = \int\limits^ .dt$

-9㏑[9.81-v/9] = t +c

㏑[9.81-v/9] = -t/9 +c

9.81-v/9 =C$e^{-\frac{t}{9} }$

-v/9 = C$e^{-\frac{t}{9} }$ - 9.81

put  v(t) = C$e^{-\frac{t}{9} }$ + 88.29

V0 =0  = C$e^{0}$ + 88.29 ⇒ C = -88.29

v(t) = 88.29 - 88.29·$e^{-\frac{t}{9} }$

Equation of motion

x(t) = $\int\limits^ . v(t)dt$ = 88.29·t + 794.6$e^{-\frac{t}{9} }$  +C₁

x = 0 c₁ = -794.6

x(t) = 88.29·t + 794.6$e^{-\frac{t}{9} }$  -794.6