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 .

The answer to the question is;

The equation of motion of the object is

x(t) = 88.29·t + 794.6  -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

=

-9㏑[9.81-v/9] = t +c

㏑[9.81-v/9] = -t/9 +c

9.81-v/9 =C

-v/9 = C - 9.81

put  v(t) = C + 88.29

V0 =0  = C + 88.29 ⇒ C = -88.29

v(t) = 88.29 - 88.29·

Equation of motion

x(t) = = 88.29·t + 794.6  +C₁

x = 0 c₁ = -794.6

x(t) = 88.29·t + 794.6  -794.6

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂