- #1

MathewsMD

- 433

- 7

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!