falling body

• Oct 25th 2010, 06:48 PM
Zamzen
falling body
A free falling body has no initial velocity. Its mass is 80kg and the air resistence is proportional to the velocity squared with the proportional constant 0.16kg/m.Let velocity be y m/s and x s. And the question is to put up a differential equation for this.

My attempt at the solution.
y(0)=0 (easy part). Now my thinking about the equation itself goes something like this.
F=ma
a=9.82
F(t) = F-Fr(t) (Fr=Air resistance)
ma(t)=ma- v^2 * 0.16(t)
divide it all by m which is 80
a(t) = 9.82 - 0.002v^2(t)
swapping out v(t) for y and a(t) for dy/dt
dy/dt=9.82 - 0.002y^2
which is correct according to the text book.
My issue is That it seems that i have solved for the accelration and not for the velocity itself. Im thankful if anyone could explain this.
• Oct 25th 2010, 10:11 PM
mr fantastic
Quote:

Originally Posted by Zamzen
A free falling body has no initial velocity. Its mass is 80kg and the air resistence is proportional to the velocity squared with the proportional constant 0.16kg/m.Let velocity be y m/s and x s. And the question is to put up a differential equation for this.

My attempt at the solution.
y(0)=0 (easy part). Now my thinking about the equation itself goes something like this.
F=ma
a=9.82
F(t) = F-Fr(t) (Fr=Air resistance)
ma(t)=ma- v^2 * 0.16(t)
divide it all by m which is 80
a(t) = 9.82 - 0.002v^2(t)
swapping out v(t) for y and a(t) for dy/dt
dy/dt=9.82 - 0.002y^2
which is correct according to the text book.
My issue is That it seems that i have solved for the accelration and not for the velocity itself. Im thankful if anyone could explain this.

Using the symbol v instead of the (stupid) symbol y:

dv/dt = 9.82 - 0.002v^2

which, perhaps more obviously now, is obviously a DE for the velocity. Solve it and you get v.