Relationship between Launch Angle & Horizontal Distance

[Anonymous_001]

Homework Statement

How to prove the correlation between launch angle (<45°) and horizontal displacement.
Independent Variable:

• Horizontal distance

Dependent Variable:

• Launch Angle

(Important) Controlled Variable:

• Firing velocity [90 m/s]

Another problem is, with the data that I have collected (e.g. angles for horizontal distances from 1-5 m), initial velocity seems to change? Or I've made an error?

2. (Useful Equations:)
I need help with the proof, perhaps something using the following equations:
x = (vi*cosθ)t
y = (vi*sinθ)t - 1/2(gt²)

With gravitational acceleration being 9.8 m/s².
Although I did not measure time (duration for each projectile), I assume this can be found using v = d/t.
This is a snippet of the data that I have collected:

 

[Ackbach]

So, to clarify: are you trying to derive a kinematics relationship theoretically? Or are you trying to make sense out of your experimental data?

 

[Anonymous_001]

So, to clarify: are you trying to derive a kinematics relationship theoretically? Or are you trying to make sense out of your experimental data?

I am more trying to make sense of my data!
I will be required to plot graphs, but I need an in-depth explanation of why I received the results that I have. I also need guidance with what would be the best to plot on a graph to justify the relationship, if possible!

 

[Ackbach]

Ah, I see. Well, the one thing I would strongly advise is to greatly increase the range of different angles over which you launch your projectile. I would probably have at least ten different angles ranging from 10 degrees to 80 degrees. The fact that you have some horizontal distances that are essentially indistinguishable from each other is a consequence of the fact that your angles are extremely close together. If you could do some video analysis frame-by-frame, that could enable you to determine if the magnitudes of your initial velocities are changing a lot. (I recommend at least 30 frames per second, and use QuickTime to analyze the video, because you can use the right and left arrow keys to forward or backward one frame at a time. Many other video players don't have that ability.)

I agree with your equations, provided the launch height and the landing height are the same, and you're ignoring air resistance. In that case, plug in zero to the $y$ equation to find out when it lands : $t=0, \dfrac{2v_i \sin(\theta)}{g},$ and it's the second solution we're after, since the first solution corresponds to the launch. Then you can plug this into the $x$ equation to obtain $x=\dfrac{2v_i^2 \sin(\theta) \cos(\theta)}{g}=\dfrac{v_i^2 \sin(2\theta)}{g}.$ So we're roughly expecting this sort of dependence.

 

[Anonymous_001]

I agree with your equations, provided the launch height and the landing height are the same, and you're ignoring air resistance. In that case, plug in zero to the $y$ equation to find out when it lands : $t=0, \dfrac{2v_i \sin(\theta)}{g},$ and it's the second solution we're after, since the first solution corresponds to the launch. Then you can plug this into the $x$ equation to obtain $x=\dfrac{2v_i^2 \sin(\theta) \cos(\theta)}{g}=\dfrac{v_i^2 \sin(2\theta)}{g}.$ So we're roughly expecting this sort of dependence.

I can kind of see what you did there, but just to clarify, how did you get 2 in the nominator, in the first equation and did y or t become 0? ($\dfrac{2v_i \sin(\theta)}{g},$) Was there a major step that you did to arrive at $\dfrac{2v_i \sin(\theta)}{g},$ ? (I'm sorry, I'm kind of slow on these things!)

 

[Ackbach]

So you set $y=0$ to obtain $0=v_i \sin(\theta) t-\dfrac{g}{2} \, t^2 = t\left(v_i \sin(\theta)-\dfrac{g}{2} \, t\right).$ Hence, we have the two solutions $t=0$ and $0=v_i \sin(\theta)-\dfrac{g}{2} \, t.$ If you solve this second equation for $t$, you will see where the $2$ originates. Does that answer your question?

 

[Anonymous_001]

So you set $y=0$ to obtain $0=v_i \sin(\theta) t-\dfrac{g}{2} \, t^2 = t\left(v_i \sin(\theta)-\dfrac{g}{2} \, t\right).$ Hence, we have the two solutions $t=0$ and $0=v_i \sin(\theta)-\dfrac{g}{2} \, t.$ If you solve this second equation for $t$, you will see where the $2$ originates. Does that answer your question?

Yes, thank you for the explanation! Using the data that I have, what graph would suit to represent it best? At first, I was considering of plotting angle [theta] against horizontal distance, but that may be too rough. Is there anything that I can plot against the horizontal distance, that can also be best compared to the equations associated with projectile motion?

 

[Ackbach]

Yes, thank you for the explanation! Using the data that I have, what graph would suit to represent it best? At first, I was considering of plotting angle [theta] against horizontal distance, but that may be too rough. Is there anything that I can plot against the horizontal distance, that can also be best compared to the equations associated with projectile motion?

Well, if you're thinking of horizontal distance as a function of the launch angle, then I would put the angle on the horizontal axis, and distance on the vertical. This is the relationship you're trying to establish, so that's the plot you need. If you plot it the other way, it's going to look like it's not a function (because it isn't!). It will fail the vertical line test, as there are generally two angles in the range of $0$ to $90$ that yield the same horizontal distance.

 

[Anonymous_001]

Well, if you're thinking of horizontal distance as a function of the launch angle, then I would put the angle on the horizontal axis, and distance on the vertical. This is the relationship you're trying to establish, so that's the plot you need. If you plot it the other way, it's going to look like it's not a function (because it isn't!). It will fail the vertical line test, as there are generally two angles in the range of $0$ to $90$ that yield the same horizontal distance.

Thank you very much for your help!

 

[Anonymous_001]

Well, if you're thinking of horizontal distance as a function of the launch angle, then I would put the angle on the horizontal axis, and distance on the vertical. This is the relationship you're trying to establish, so that's the plot you need. If you plot it the other way, it's going to look like it's not a function (because it isn't!). It will fail the vertical line test, as there are generally two angles in the range of $0$ to $90$ that yield the same horizontal distance.

Another question, is there a good way to find the initial velocity? Consider the example of something fired from a cannon. I'd imagine, the more elevated the cannon is, the more initial velocity is needed to fire the ball at that angle. What do you do about cases like this, especially when you have a formula that connects horizontal displacement with the angle fired at?

 

[Ackbach]

Well, there are several ways you could get the initial velocity:

1. If your experimental gear allows you to set it, you can simply accept that value as correct.
2. You could measure it explicitly by using video analysis frame-by-frame.
3. You could measure it implicitly if you know the horizontal distance traveled and the launch angle by solving the above equation to yield $v_i=\sqrt{\dfrac{xg}{\sin(2\theta)}}.$

 

[Anonymous_001]

Well, there are several ways you could get the initial velocity:

1. If your experimental gear allows you to set it, you can simply accept that value as correct.
2. You could measure it explicitly by using video analysis frame-by-frame.
3. You could measure it implicitly if you know the horizontal distance traveled and the launch angle by solving the above equation to yield $v_i=\sqrt{\dfrac{xg}{\sin(2\theta)}}.$

I just plotted the graph of horizontal distance against angle and vice versa. What should I do, if the line does not go through the origin, thus showing that it is not linear, as it is supposed to be?

 

[Ackbach]

Well, the relationship is not linear in the first place. It should go through the origin, though, or close to it. Can you post your graph?

 

[Anonymous_001]

Well, the relationship is not linear in the first place. It should go through the origin, though, or close to it. Can you post your graph?

When connecting from dot-to-dot and not using linear fit, the arrangement of points looks something like https://image.prntscr.com/image/GCahGEWMTt2YMqJPeT-OQw.png. One of the points is (0,1). My min. and max. line do not intersect with the original line at the same point, thus making the graph look chaotic.
That is why I was unsure, if there was anything that I could do with the angle to make it go through closer to the origin. I thought of using something like, sin(theta)*cos(theta) and plotting that against the horizontal displacement.

 

[haruspex]

Independent Variable:

• Horizontal distance

Dependent Variable:

• Launch Angle

That is backwards. The angle is the independent variable, the range depends on it.

The data you posted do not make much sense. All of the distances were either 2m or 3m? Surely you measured them more precisely than that!
And did you really not try an angle greater than 6 degrees?

 

[Ackbach]

Yes, yes: to haruspex you listen. Part of your problem is that your graph has very little perspective, because your angle range is so small. If you were to graph your experimental results for $10$ degrees through $80$ degrees, say, I think you'd have a more complete picture.