I found a mechanics question in the 2022 Mathematics Extension 2 HSC (the final exam in Australia) whose solution didn't quite make sense to me. This is the question:
My initial thought was to create two differential equations, one for the travel upwards where: acceleration = -g - 0.1v and one for the downward travel where: acceleration = -g + 0.1v
This is because when travelling up, the air resistance forces the projectile in the down direction, but when travelling down, the air resistance forces the projectile in the downward direction. However, I couldn't end up getting a solution and I just got a big messy system of equations I couldn't be bothered to solve (I probably couldn't even solve it if I wanted to).
However, the solution's official of question didn't do what I did and just used the equation: acceleration = -g - 0.1v
Their solution is too long to put in an image, so I'll give the link to it. The question is Q16(b): https://www.educationstandards.nsw.edu.au/wps/wcm/connect/85d54825-e505-4f7c-ad7b-48d3c4aeee5d/2022-hsc-maths-ext-2-mg.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-85d54825-e505-4f7c-ad7b-48d3c4aeee5d-ok9CW4N
Also, if you want the link to the exam itself, its here: https://www.educationstandards.nsw.edu.au/wps/wcm/connect/b1c14c48-d8e7-4b10-9f28-84ce9f0d65a8/2022-hsc-mathematics-extension-2.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-b1c14c48-d8e7-4b10-9f28-84ce9f0d65a8-og8aQIY
I find it hard to believe that such a massive exam (its the final exam for the final year of highschool) could have an incorrect solution, but I also don't see how this could be correct since the equation they use doesn't work for the travel back down. Thanks to anyone who can hep :)
When dealing with linear drag you do not need to change the sign on your velocity since the sign of the velocity "works itself out" since it is negative going down and positive going up. If you were dealing with quadratic drag with a $v^2$ term then you would need to split the upward and downward motion into two separate equations since the drag acceleration needs to oppose the current motion but $v^2$ will always be nonnegative.