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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:

enter image description here

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.

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  • $\begingroup$ Have you tried plotting on desmos? The answer seems quite apparent immediately doing that ;) $\endgroup$
    – Thormund
    Commented Jun 4, 2020 at 0:16
  • $\begingroup$ @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? $\endgroup$
    – abu barley
    Commented Jun 4, 2020 at 0:22
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    $\begingroup$ @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. $\endgroup$
    – abu barley
    Commented Jun 4, 2020 at 0:27

1 Answer 1

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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.

enter image description here

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  • $\begingroup$ Thank you for the help. I'm still confused, because if I choose to apply the professor's approach to the second problem in my post, I get a negative value for time. I'm not sure if I agree with your assertion that the "sign is automatically built into the definition of y." From what I understand, the formula we're using gives us a displacement, and it just so happens that displacement often (but not always) = the y position. $\endgroup$
    – abu barley
    Commented Jun 4, 2020 at 0:49
  • $\begingroup$ @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 $y$. $\endgroup$
    – Vincent Thacker
    Commented Jun 4, 2020 at 0:54
  • $\begingroup$ Ah, then can you help me understand why the book approaches the problem with a "-y"? I will edit my question by inserting the picture from my book. $\endgroup$
    – abu barley
    Commented Jun 4, 2020 at 0:57
  • $\begingroup$ @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 $80$. $\endgroup$
    – Vincent Thacker
    Commented Jun 4, 2020 at 1:00
  • $\begingroup$ Good catch, but the time is negative even when using 80. I'll edit my question again with the proper equations :) $\endgroup$
    – abu barley
    Commented Jun 4, 2020 at 1:05

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