If we throw something upwards with some initial velocity (of course), then it's the same whether the acceleration upward is considered positive or negative, right?
But by convention, the upward direction is considered positive. That's good, but see-
In that case, if initial velocity is 20m/s and g=10m/s², then the displacement upward in 2 seconds would be 40m if acceleration due of gravity is positive upward and the displacement would be 10m if g is negative upwards.
So it matters right?
I am not getting the reason what the convention means to say actually..
Best Answer
Suppose we adopt the convention that a distance upwards is positive and a distance downwards is negative. Velocity is given by:
$$ v = \frac{dx}{dt} $$
So if the object is moving upwards its position increases, i.e. gets more positive with increasing time so $dx \gt 0$ and $dt \gt 0$. That means an object moving upwards has a positive velocity. The same argument tells us that an object moving downwards has $dx \lt 0$ and therefore it has a negative velocity.
So by choosing the sign convention for the distance we automatically get a sign convention for the velocity.
But acceleration is given by:
$$ a = \frac{dv}{dt} $$
So now we have a sign convention for velocity this also defines the sign convention for acceleration. If something is accelerating upwards it has $dv \gt 0$ and therefore a positive acceleration. Likewise something accelerating downwards has a negative acceleration.
In your question you've used the usual convention that distances up are positive, so the initial velocity of $+20$ m/s means the object is moving upwards. And since the gravitational acceleration is downwards we have to write $g = -9.81$ m/s$^2$.
All you have to do is feed these values with their signs into your equations of motion and you will get the correct answer.
You do often see the gravitational acceleration referred to as just $9.81$ m/s$^2$ i.e. without a sign. This is lazy terminology, and it means that the magnitude of $g$ is $9.81$ m/s$^2$.