# Correct sign of vertical displacement in projectile motion

I understand how to do this problem perfectly fine.

I am posting here however because I have a disagreement with my professor and classmates in finding the final y-coordinate of the projectile.

I am confident that to find the final y-coordinate of the projectile, the correct equation should be:

$$-y = (100sin60^o )t + 0.5 (-32.2)t^2$$

My professor, however, says that my equation for finding the final y-coordinate of the projectile is categorically incorrect. It should be:

$$y = (100sin60^o )t + 0.5 (-32.2)t^2$$

So, who's right?

I believe I'm correct because the formula we're using here is the formula for displacement. Displacement is a vector, and it's the change in position. The final y-position of the projectile is obviously "-y" and the initial y-position of the projectile is 0. Therefore, the vertical displacement should be "-y," not just "y."

In addition, there is a very similar problem in the text. In the below problem, if we solve it with the professor's method/equation (in addition to the other requisite equations):

$$y = (80sin30^o )t + 0.5 (-32.2)t^2$$

$$x = (80cos30^o)t$$

$$y = -0.04x^2$$

We would actually end up with a time of flight that is negative:

If we solve it with my method:

$$-y = (80sin30^o )t + 0.5 (-32.2)t^2$$

$$x = (80cos30^o)t$$

$$y = -0.04x^2$$

We get the correct answer in the back of the textbook, and a time of flight that is positive.

• Have you tried plotting on desmos? The answer seems quite apparent immediately doing that ;) Commented Jun 4, 2020 at 0:16
• @Thormund--I don't have graphing software. I see that Desmos is free but I'm unfamiliar with it. Who's correct, me or the prof? Commented Jun 4, 2020 at 0:22
• @Thormund--I'm not asking about a specific computation. There's not a single number in my question apart from the ones contained within formulas. I asked a conceptual question about the proper application of a formula. The "homework problems" contained within my question are for illustrative purposes. I could have asked the same thing without the homework problems, but it would have needlessly complicated the question. Commented Jun 4, 2020 at 0:27

You are confusing distance with coordinates. The correct equation to use would be $$y$$. Why? Because the variable $$y$$ itself can be either positive or negative, and this sign is automatically built into the definition of $$y$$. It would be extraneous (and incorrect) to add an extra negative sign in front of it.
The same thing applies if you choose to orient the $$y$$-axis downwards. In this case, the value of $$g$$ will be positive as it points in the same direction as $$y$$. The negative sign in the quadratic equation of the surface will also need to be removed to reflect this.
• @abubarley "displacement often (but not always) = the y position": That is not correct. The $y$ value is the vertical component of the displacement at all times. Regardless of whether it is negative, it is still $y$. Commented Jun 4, 2020 at 0:54
• @abubarley If you are getting a negative value for $t$, it means that the projectile will never hit the slope. The parabola of the slope is entirely inside that of the projectile. This is probably because you used $100$ instead of the given value in the question, which is $80$. Commented Jun 4, 2020 at 1:00