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# Differential Equation watch

1. A question in a book about ordinary differential equations states:

"An object of mass m is thrown vertically upwards. The air resistance is proportional to the square of the velocity. Derive the differential equation relating the velocity v and time."

When I drew a diagram of a ball being thrown upwards I drew onto it an arrow pointing downwards marked mg.

Now there would also have to be another arrow marked on the diagram to indicate air resistance and I would imagine this would also point downwards.

In the book the answer is given as m(dv/dt) + kv^2 = mg

I know they have first calculated the net forces acting on the body, and then used F=ma

Does anyone know how they got this asnwer? Thanks very much!
2. (Original post by little pixie)
In the book the answer is given as m(dv/dt) + kv^2 = mg

I know they have first calculated the net forces acting on the body, and then used F=ma
That's pretty much all you need to do. If you're not getting this answer, show some working.
3. The book's wrong. kv^2 and mg should have the same sign. It should read m(dv/dt) + kv^2 + mg = 0
4. (Original post by spex)
The book's wrong. kv^2 and mg should have the same sign. It should read m(dv/dt) + kv^2 + mg = 0
It's okay either way, k could be positive or negative. But yeh, you'd normally write it like that.

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