Question

• Suppose an object falls from a great height on a planet where the acceleration constant of gravity is g = 7.29. Suppose that the resistance of the atmosphere is proportional to the square of the speed of the object with constant of proportionality k = 0.36. Establish and solve an Initial Value Problem to express the velocity of the object as a function of time. Find the terminal velocity of the object. Graph this function. Then express the fallen distance as a function of time. Graph this function. Suggestion:
• 
  g- ku? = (va- vku) (vg+ vku)
  

Answer 1

cit = st 8- kr? Here og - 1 dv= Suppose an object falls from a great height on a planet where the acceleration constant of gravity is g = 7.29. Suppose that the resistance of the atmosphere is proportional to the square of the speed of the object with constant of proportionality k = 0.36. Establish and solve an Initial Value Problem to express the velocity of the object as a function of time. Find the terminal velocity of the object. Graph this function. Then express the fallen distance as a function of time. Graph this function. Resultant accelerodsm acting on the body: dv 8-kv2, where v is the velocity Now, dr du Cog-Je v) (v +1 kv) t dt 2.88 188-kv sq +5kv 17.29 = 217 and rk=√0.36= 0.6 So, we have 5.4 (2.7-6.6v) (2.7+0.60) On integration, we get 2.7+0.6V 27-0-6v ne 3.At ce where is constant of interprasin c=1 1.2 V By componendol e 3+2-4 tl Aclividendo 5.4 83.24€ zrite -3,24ti 3.244 e --V=4.5 1-e -3.24+ ite + 1 Now, terminal relocity vt $: Lim 3.745 4. = 4.5 -3.202 Ite -3.24t e 400 3.24+ 2.7+ 0.6v 2.7-oibr ,ا AŁ t=0, V=0 tl = 4.51 32A Vq=ttoo st So, velveity: V(t) = 4.5 2.245 +} & Ve (terminal vedocity) = 4.5