# Projectile Motion: two positive times?

1. Sep 10, 2010

### Living_Dog

I explained to my class that in projectile motion one always chooses the positive time answer with the negative time answer being dropped. Then a student asked if two positive time answers were possible. I just looked at it assuming that t+/- are both positive and got the result that for t+ to be positive, then

yf > yi.​

But for t- to be positive (for the same problem)

yi > yf.​

I believe this is a contradiction and so the answer to the student's question is 'no' - two positive time answers, in the same problem, are not possible. I need confirmation of this before I give the answer to the class.

-joe

2. Sep 10, 2010

### Staff: Mentor

I don't understand the context. Why can't you have a projectile motion question with two answers, both at positive times? Throw a ball in the air. At what time(s) will it be 10 ft above the ground?

3. Sep 10, 2010

### Living_Dog

If the initial and final positions are fixed, then solving for the time of flight should yield only one positive time.

yf = yi + voy*t - 4.9*t2

The solution of this is the quadratic formula:

$$t = \frac{-v_{oy} \pm \sqrt{v_{oy}^{2} + 19.6*\Delta y}}{-g}$$​

There are two answers: t+ and t-.

For both to be positive, one sets up the inequality for each and deduces the result above, namely:

For t+ > 0 this implies Dy > 0; and
for t- > 0 this implies Dy < 0.

4. Sep 11, 2010

### Staff: Mentor

I don't understand how you are drawing this conclusion. If Yf > Yi, then both times will be positive if Vo is positive; If Yf < Yi, then one of the times will be negative.

5. Sep 12, 2010

### Living_Dog

Using the quadratic solution for the time (link is to the above formula):

https://www.physicsforums.com/latex_images/28/2874555-0.png [Broken]

I set t+ > 0 and t_ > 0 and look for the necessary condition to ensure this.

For t+ > 0, yf > yi, and

for t_ > 0, yi > yf.

But this was my original question, was my deduction correct?

Last edited by a moderator: May 4, 2017
6. Sep 12, 2010

### Staff: Mentor

How can your deduction be correct? You claim that you cannot have a case where you have two positive time solutions and I gave a trivial example that shows you can! Here's one with numbers: Throw a ball upward with initial velocity = +10 m/s. At what time(s) will it be 5 m above the starting position?

Show the details of how you made your deductions--I suspect either a mathematical error or that I am misinterpreting the problem you are trying to solve.

7. Sep 12, 2010

### Living_Dog

Ok, here goes...

$$t_{+} = \frac{v_{oy}}{g} + \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} > 0$$

Canceling the common denominator, g, and bringing the voy term to the other side then I get:

$$\sqrt{v_{oy}^2 + 19.6\Delta y} > -v_{oy}$$

Squaring both sides I get:

$$v_{oy}^2 + 19.6\Delta y > v_{oy}^2$$

Now the initial y-velocity component can be canceled, which yields:

$$19.6\Delta y > 0$$

Dividing out the constant factor of 19.6,

$$\Delta y > 0$$

which yields the result for t+, namely:

yf > yi.

--

For t- the result is:

$$t_{-} = \frac{v_{oy}}{g} - \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} > 0$$

Canceling the common denominator, g, and bringing the square-root to the other side then I get:

$$v_{oy} > - \sqrt{v_{oy}^2 + 19.6\Delta y}$$

Squaring both sides, and canceling the voy term leaves:

$$0 > 19.6\Delta y$$

Dividing out the constant factor of 19.6,

$$0 > \Delta y$$

which yields the result for t-, namely:

yi > yf.

--

I see something now I didn't see before - that in one attempt I moved the voy term, and in the other I moved the square-root term. Thus I obtained two different results. However, this isn't wrong, and your previous, intuitive, answer is correct.

So what's going on here???

Thanks for your help Doc Al.

-joe

8. Sep 12, 2010

### Staff: Mentor

I see the problem. You made a sign error with respect to Δy:

This:
yf = yi + voy*t - 4.9*t2

becomes:
0 = -Δy + voy*t - 4.9*t2

9. Sep 12, 2010

### Living_Dog

That minus sign is common to both t+ and t-. The only thing that changes is that the results are switched. Thanks for that. But I still think that something is weird about the "square both sides of the inequality" step.

-joe

10. Sep 12, 2010

### Andy Resnick

It's definitely possible to have two positive times- for example, launch two projectiles at two angles, each equidistant from 45 degrees. Both projectiles will hit the target, but take 2 different paths (and 2 different times).

Does that help?

11. Sep 12, 2010

### Staff: Mentor

Good point. Putting the sign of Δy aside for the moment: Yes, the problem is where you square both sides. Not valid!

Example:

2 > -3

Square both sides:

4 > 9 :yuck:

Sorry about not spotting that earlier! :uhh:

Starting with this:
Clearly the inequality will be satisfied for any value of Δy that keeps the discriminant positive. (If it's not positive, there's no real solution.)