Question

I'm using a launching mechanism at the angles(in degrees) of 10, 20, and 30. I launch at these angles using the mechanism and measure where it lands in meters. I also account for the amount of clicks the launching mechanism uses to launch the object. (1,2 or 3 clicks available for each angle) How do I find my initial velocity using this information?

Answer 1

Short(1 click of the launching mechanism):
{10 degrees- 149.5cm, 20 degrees- 162cm, 30 degrees- 171.5cm}
Medium(2 clicks of the launching mechanism):
{10 degrees- 205cm, 20 degrees- 231.5cm, 30 degrees- 250cm}
Long(3 clicks of the launching mechanism):
{10 degrees- 286cm, 20 degrees- 338.5cm, 30 degrees- 380.5}

Those are my measurements of where the object fired lands on the ground.

Answer 2

If you'd like me to do the differential equations derivation, I can do so. But if you're just looking for an equation, you can use the following:\[v_{i}=\sqrt{\frac{gx}{\sin(2\theta)}}\]'g' is gravitational acceleration, theta is angle, x is the range

Answer 3

and having x as my range refers to my lengths recorded. I unfortunately have never had any experience with differential equations, so if you'd like to teach me, that would be very appreciated.

Answer 4

Oh, well I won't use that method then because I doubt you're held accountable for something you don't know about. You should be familiar with the kinematic equations though. You can start with those and then combine a couple to derive the expression above. I'll go ahead and type it out, but I promise you at the minimum the expression above is correct for this application.

Answer 5

This is a lab experiment for my physics course; for the lab, we have to launch a steel ball from a launching mechanism mentioned about and those are my measurements. Once I've measured these, I need to find an accurate point in which the ball lands on a paper bulls eye that I have laid over the recorded measurements. From there I have to determine the final velocity (including x and y components then take the necessary measurements and calculate where the bulls eye will hit for the final trial run.
I know that's very lengthy, but do you have any suggestions as to how I would go about that process? I spent two hours trying to figure it out to no avail.

Answer 6

I'm not sure I quite understand your description of the procedure.

So you've already measured 3 angles, each one at 3 different initial velocities.

I'm confused about the bullseye your laying over these. Also, by final velocity (and its components) do you mean initial velocity? Or is this a completely different experiment?

Answer 7

Yes, I have three different angles that I launch from, each time I launch, I launch it with one, two or three clicks of the mechanism, thereby increasing the speed and distance of the launch at the angles.
I have to lay the bullseye on the floor where I have measured the ball to land at each angle that they were launched and hope that the ball lands on the bullseye. The final velocity, meaning, the same experiment, but I assume the final velocity will be different than the initial by the different launch angles.

Answer 8

Ahh okay, so you've done some measurements and now you'd be able to predict where the ball will be, got it.

Final velocity will only be different than initial velocity if there is an elevation difference. If you launched from the ground and the ball landed on the ground, then there shouldn't be a difference.

Answer 9

Yes, alright, I just wanted to make sure that i didn't have to find a whole other velocity. I think I understand, I understand from here on out. I won't need to use those predictions to find anything else?

Answer 10

I don't think so. I just wanted to double-check that the launch height and landing height were equal. That is the assumption of the expression I passed along to you! But so long as you can calculate initial velocities from the above data using the expression, you should have almost any values necessary to predict.

Answer 11

Thank you, I really appreciate you helping me. This is my first lab where I needed to create my own steps and procedures, I was lost on the necessary calculations. Thank you again!

Answer 12

Not a problem. If more questions come up later, just continue in this thread. As promised, let me quickly derive the expression I gave to you so that you can see where it comes from:

We start with two kinematic equations, one for horizontal motion and one for vertical motion:

Horizontal:\[x=v_{o}\cos(\theta)t\]Vertical:\[y=y_{o}+v_{0}\sin(\theta) t-\frac{1}{2}gt^{2}\]
We take y=y_0=0 and then solve the vertical equation for time 't' to find:\[t=\frac{2v_{0}\sin(\theta)}{g}\]
Now, time should be the same for both horizontal and vertical motion (obviously this is true, as the both are a part of the same projectile flight). We substitute this expression for time into the horizontal equation:\[x=v_{0}\cos(\theta)(\frac{2v_{0}sin(\theta)}{g})\]
We simplify by using the identity:\[2\sin(\theta)\cos(\theta)=\sin(2\theta)\]
This gives us our final expression for initial velocity:\[v_{0}=\sqrt{\frac{gx}{\sin(2\theta)}}\]