What is causing confusion in solving for projectile motion?

[alexcroteau]

Hello, I am a first year science teacher doing my best with teaching physics for the first time (my degree is chemistry but I am in a very small school).

I am teaching projectile motion. I was creating a worksheet and trying to solve a problem I made up when I realized something wasn't working out.

I made a problem where an object is launched with a known initial velocity, a known time of flight, and a known distance to the target (target is at same height as launch). I wanted students to find the angle of launch.

When I started to solve I used Vox = (Xf-Xo)/t to find the x component of the initial velocity, then I can simply solve for the angle using trigonometry. All of this checks out and works fine.

However, I wanted to see if I got the same angle if I solved for the y component instead of the x component. I used (yf-yo) = Voy*t + 1/2at^2

The problem is that the y component I get from solving this is always half of the y component I got from trigonometry using the x component I solved for earlier. Thus my angle is also wrong. This doesn't make sense to me. Hopefully someone can help.

Sorry this was a bit wordy. I can add values if it would help.

 

[berkeman]

Hello, I am a first year science teacher doing my best with teaching physics for the first time (my degree is chemistry but I am in a very small school).

I am teaching projectile motion. I was creating a worksheet and trying to solve a problem I made up when I realized something wasn't working out.

I made a problem where an object is launched with a known initial velocity, a known time of flight, and a known distance to the target (target is at same height as launch). I wanted students to find the angle of launch.

When I started to solve I used Vox = (Xf-Xo)/t to find the x component of the initial velocity, then I can simply solve for the angle using trigonometry. All of this checks out and works fine.

However, I wanted to see if I got the same angle if I solved for the y component instead of the x component. I used (yf-yo) = Voy*t + 1/2at^2

The problem is that the y component I get from solving this is always half of the y component I got from trigonometry using the x component I solved for earlier. Thus my angle is also wrong. This doesn't make sense to me. Hopefully someone can help.

Sorry this was a bit wordy. I can add values if it would help.

Welcome to the PF.

Problems like this one can sometimes have two answers, since parabolic motion is involved. Can you post the problem and your work? We can see if there are two solutions to the problem.

 

[alexcroteau]

I set it up like so:

A projectile is fired with an initial velocity of 25.0 m/s at an unknown angle. The projectile flew through the air for 2.0 seconds before hitting a target placed 33.0 meters away at the same height as the launch height. Solve for the unknown angle.

As I said in the original post I can solve for the angle no problem using the x component.

Vox = (Xf-Xo)/t
Vox = 33.0/2
Vox = 16.5 m/s

cos(theta) = 16.5/25.0
theta = 48.7°

The third side of this right triangle (Voy) should be 16.5^2 + b^2 = 25.0^2 b = 18.8 = Voy

Here is where I have the problem, with the y component

(Yf-Yo) = Voy*t + 1/2at^2

0 = Voy*2 + 1/2 (-9.81)* 2^2

0 = 2Voy - 19.62

19.62/2 = Voy

Voy = 9.81

I think my problem is that I am not actually solving for the y component. I simply can't figure out why. Nor can I figure out what I am solving for, since logically it should just be reverse process of solving this type of problem.

 

[Nathanael]

You've given too much information. Give any two of the three variables given (initial speed, distance, and time of flight) and the problem will be solvable. Therefore the third piece of information can not be chosen arbitrarily; it must be made consistent with the other two variables.There are three combinations you can choose from:If you give initial speed, V₀, and time of flight, Δt, then the solution is:
$g\Delta t = \Delta V_y = 2V_0\sin\theta \Rightarrow \theta = \arcsin(\frac{g\Delta t}{2V_0})$

If you give the horizontal displacement, Δx, and the time of flight, Δt, then the solution is:
$\Delta x = V_x\Delta t=V_0\cos(\theta)\Delta t \Rightarrow \theta = \arccos(\frac{\Delta x}{V_0\Delta t})$

If you give the initial speed, V₀, and the horizontal displacement, Δx, then the solution is:
$\Delta t = \frac{\Delta V_y}{g} = \frac{2V_0 \sin\theta}{g}$
$\Delta x = V_x\Delta t = V_0\cos\theta\frac{2V_0 \sin\theta}{g}=\frac{2V_0^2}{g}\tan\theta \Rightarrow \cos(\theta)\sin(\theta) = \frac{g\Delta x}{2V_0^2}$Notice in this third combination, there will usually be two solutions (or else there is either no solution, or θ is 45°) which is what Berkeman was talking about. That isn't the problem though; the problem is that you have not chosen the three variables so that the different methods give the same answer (in other words your excess information is inconsistent).

 

[alexcroteau]

That is what is bothering me. I solved the following problem first and then used those values to create the problem I wrote above.

A projectile is fired with an initial velocity of 25.0 m/s at an angle of 48 degrees. How long did the projectile fly through the air before hitting a target placed some distance away at the same height as launch height? How far away is the target?

25.0 sin (48) = Voy = 18.58 m/s
25.0 cos (48) = Vox = 16.73 m/s

(Yf-Yo) = Voy*t + 1/2at^2

0 = 18.58*t + 1/2(-9.81)(t)^2
t = 0 or t = 3.8s <----- Wow there it is. I messed up right here when I solved my original problem. The first time I did not get 3.8 which made me set the entire other problem up incorrectly.

Disregard everything I have said to this point, I simply had a calculation error in my initial problem, causing the inconsistency you mentioned in my variables.

When I fix the problem to read:

A projectile is fired with an initial velocity of 25.0 m/s at an unknown angle. The projectile flew through the air for 3.8 seconds before hitting a target placed 63.4 meters away at the same height as the launch height. Solve for the unknown angle.

It fixes the entire issue. I'm so glad that's over. Thanks

 

[ogg]

My formal background is also in chemistry, but with zero (or less) teaching experience. I'd strongly suggest that you always always use a spreadsheet program (ie Excel) to set up and solve any numerical problem you create. Of course you'll have your own method of doing this, but I'd suggest that you use separate (adjacent) rows for variables (your Yf should have been Ymax, right? or did I misunderstand that Ymax occurred at ½t?), a second for numerical values, and a third for units.
v = Vo + a*t
25=? + 0*3.8
m/s = m/s + m/s² * s
(where ' becomes indispensable for entering bare =,+,- symbols in cells) (ie |25 | '= | ? | '+ | 0 | * | 3.8| and a cell with the formula =ref1+ref2*ref3 where refx is cell reference and | indicates cell boundary.)
You should be able to eliminate numerical /calculational errors this way, and once you've got a system, reduce your logical errors (and make them explicit, one hopes).
I've been using Excel since ...the early 1980's (almost certainly before you were born) and it is a bit slower that a calculator+pen+paper but allows you to notate your work. This means that these workbooks can become archival and can be referred to and reused days, weeks, months, years after they were created (and with simple change in numerical values, reused). I'd suggest one problem per tab, but that's another personal choice.
...I wonder if there would be any educational value in providing "the answers" to the students as excel files ...?
There is several methods to display cell formulas in Excel (as well as tracing cell dependencies). I frequently use the F2 key to color code the formulas to check them, FWIW. (Adding the trace... buttons to the top ribbon (as well as the erase...) is recommended.