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Question

We can take the kinematic equations for constant acceleration and apply them to projectiles in free fall. One of the equations then reads:

$$\Delta y=(v_0\sin \theta )t-\frac{1}{2}gt^2$$

where $v_0$ is the projectile's initial velocity, $\theta$ is the angle of projection, $g$ is the gravitational field strength, and $t$ is the time taken to reach some vertical displacement $\Delta y$.

Rearranging into a quadratic function gives:

$$\frac{1}{2}gt^2-(v_0\sin \theta)t+\Delta y=0$$

and this has the roots:

$$t=\frac{v_0 \sin \theta ±\sqrt{(v_0\sin\theta)^2-2g\Delta y}}{g}$$

I have always thought that there are two roots because any projectile reaches the displacement $\Delta y$ at two moments: once while ascending and once while descending. Is this the correct way to interpret the roots? Does this physical interpretation always hold true, or are there exceptions?

Motivation

A colleague and I both solved this projectile question by Walter Lewin. I used the "+" when dealing with the two roots, and he used the "-". We both acquired the same result ($t=1.549\text{ s}$), but using two different methods. I don't want to post all of the details of our solutions because it would be too lengthy. I'm trying to figure out the mistake in my work or in my colleague's work, and our two approaches are mainly distinguished by how we dealt with the two roots for $t$.

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marked as duplicate by sammy gerbil, Jon Custer, glS, Bill N, rob Jul 20 '18 at 20:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For anyone interested, my colleague essentially solved the quadratic twice: once using the conditions x = 10, y > 8 and then a second time using the conditions x = 15, y = 8. (Before solving for the roots, he did some substitution using other equations, e.g., $vtcosθ = x$) Then he found where the two lines intersected to get the minimum time. But when solving for the roots he used the "-", arguing that we seek the smallest time. $\endgroup$ – jdphysics Jun 22 '18 at 15:37
  • $\begingroup$ Possible duplicate of What do negative times and negative distances represent in parametric projectile equations? $\endgroup$ – sammy gerbil Jun 22 '18 at 15:43
  • $\begingroup$ @sammygerbil, thanks for the link. I've removed that question about negative time to focus my question only one what isn't found elsewhere on the site. $\endgroup$ – jdphysics Jun 22 '18 at 15:45
  • $\begingroup$ If you both got the same answer by different methods, why do you think one of you made a mistake? Problems can often be solved by different methods. If they make the same assumptions they always lead to the same answer. $\endgroup$ – sammy gerbil Jun 22 '18 at 15:45
  • $\begingroup$ @sammygerbil, you've hit the nail on the head--this is exactly why I'm asking the question. If neither of us made a mistake, then my understanding about how to interpret the two roots is flawed. $\endgroup$ – jdphysics Jun 22 '18 at 15:47
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I have always thought that there are two roots because any projectile reaches the displacement $\Delta y$ at two moments: once while ascending and once while descending. Is this the correct way to interpret the roots? Does this physical interpretation always hold true, or are there exceptions?

Yes. This is what the two roots represent. There's obviously trouble when they meet and vanish (i.e. you're asking for a $\Delta y$ that's too large for the initial velocity) but so long as they're real, they must have those characteristics.

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