Projectile Motion: two positive times?

[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

y_f > y_i.​

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

y_i > y_f.​

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.

Thanks in advance for any help you may give me.

-joe

 

[Doc Al]

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?

 

[Living_Dog]

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

y_f = y_i + v_oy*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.

Thus the contradiction.

 

[Doc Al]

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.

 

[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

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

For t₊ > 0, y_f > y_i, and

for t_{_} > 0, y_i > y_f.


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

 

[Doc Al]

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

For t₊ > 0, y_f > y_i, and

for t_{_} > 0, y_i > y_f.


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

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.

 

[Living_Dog]

...
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.

Ok, here goes...

t_{+} = \frac{v_{oy}}{g} + \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} &gt; 0
​

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

\sqrt{v_{oy}^2 + 19.6\Delta y} &gt; -v_{oy}
​

Squaring both sides I get:

v_{oy}^2 + 19.6\Delta y &gt; v_{oy}^2
​

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

19.6\Delta y &gt; 0
​

Dividing out the constant factor of 19.6,

\Delta y &gt; 0
​

which yields the result for t₊, namely:

y_f > y_i.

--

For t_- the result is:

t_{-} = \frac{v_{oy}}{g} - \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} &gt; 0
​

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

v_{oy} &gt; - \sqrt{v_{oy}^2 + 19.6\Delta y}
​

Squaring both sides, and canceling the v_oy term leaves:

0 &gt; 19.6\Delta y
​

Dividing out the constant factor of 19.6,

0 &gt; \Delta y
​

which yields the result for t_-, namely:

y_i > y_f.

--

I see something now I didn't see before - that in one attempt I moved the v_oy 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

 

[Doc Al]

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

y_f = y_i + v_oy*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}​

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

This:

y_f = y_i + v_oy*t - 4.9*t²​

becomes:

0 = -Δy + v_oy*t - 4.9*t²​

Now apply the quadratic formula.

 

[Living_Dog]

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

This:

y_f = y_i + v_oy*t - 4.9*t²​

becomes:

0 = -Δy + v_oy*t - 4.9*t²​

Now apply the quadratic formula.

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.

Thanks for all your help.


-joe

 

[Andy Resnick]

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 [...]

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?

 

[Doc Al]

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.

Thanks for all your help.


-joe

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


Sorry about not spotting that earlier!

Starting with this:

\sqrt{v_{oy}^2 + 19.6\Delta y} &gt; -v_{oy}
​

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.)