 #1
 433
 7
Main Question or Discussion Point
I've attached the question along with its solution. It seems fairly simple yet the answer doesn't seem quite right to me.
It appears as though they let downwards be + and that ## m\frac{dv}{dt} = mg + 0.4v## where v is negative since the ball is on its way up and there is a + instead of a  on the air resistance since the direction of motion is up but the air resistance is opposite (same direction as the positive axis), so then the equation becomes: [1] ## m\frac{dv}{dt} = mg  0.4v_s## and ##v_s## is simply the speed. This makes sense, but when I switch the case and let upwards be +, then I have: ## m\frac{dv}{dt} = mg  0.4v## and if I let ##v_s## = speed, then [2] ## m\frac{dv}{dt} = mg  0.4v_s##. Given the only difference is a change of assigning a direction as positive, shouldn't [1] and [2] be equivalent? Am I missing something here? It seems as though assigning downwards as + yields a different solution than if upwards is assigned as +...
Any clarifications would be greatly appreciated!
It appears as though they let downwards be + and that ## m\frac{dv}{dt} = mg + 0.4v## where v is negative since the ball is on its way up and there is a + instead of a  on the air resistance since the direction of motion is up but the air resistance is opposite (same direction as the positive axis), so then the equation becomes: [1] ## m\frac{dv}{dt} = mg  0.4v_s## and ##v_s## is simply the speed. This makes sense, but when I switch the case and let upwards be +, then I have: ## m\frac{dv}{dt} = mg  0.4v## and if I let ##v_s## = speed, then [2] ## m\frac{dv}{dt} = mg  0.4v_s##. Given the only difference is a change of assigning a direction as positive, shouldn't [1] and [2] be equivalent? Am I missing something here? It seems as though assigning downwards as + yields a different solution than if upwards is assigned as +...
Any clarifications would be greatly appreciated!
Attachments

70.1 KB Views: 363