Recent content by eddywalrus

Thank you for your contribution, but it doesn't really satisfy my curiosity :( In the document it refers to angles above the horizontal: Thank you for your help though!

Thank you for your response. But isn't it the other way round; at 10 degrees (i.e. 10 degrees upwards from the horizontal) most of the velocity is in the horizontal (x), rather than the vertical (y), right?

But is there an explanation for why it is the way it is? The fact that the impact is zero at 0 and 90 degrees only indicates that the impact increases with increasing angle, but doesn't explain it per say. Thanks for your help anyway!

Could you explain this a bit further please?

But the fact that the impact is zero at 0 and 90 degrees, and non-zero in between, only suggests that, as you said, the impact of air resistance on the range must increase and get back to zero -- it doesn't explain why, for example, the impact of air resistance on range is greater for 80 degrees...

Nope, as in why the impact of air resistance on range increases as the angle increases; why is it that, at higher angles, the range is reduced by so much more than it is at lower angles? Thanks for your help!

Thanks for the response and clarification, but would you be able to provide an explanation for why this is the case (if the angle x is restricted to 0 < x < 90)?

I recently read that the impact of air resistance on the horizontal range increases as the launch angle increases (http://moodle.davidson.edu/moodle2/pluginfile.php/121168/mod_resource/content/2/Brancozio%20fly-ball%20paper.pdf). A graph depicting this is attached. Is there a reason for why this...

Thank you very much for your explanation -- I understand it now!

Thank you for your help, but I think you misunderstood my question -- I probably should have made it clearer. My bad, sorry. I get how you would derive force = (change in momentum)/(time), but I'm unsure of why "time" in this instance is the time it takes for the particle to travel to the other...

Here is a screenshot from a page from a textbook that explains how to derive the ideal gas law: In the third bold line, I don't understand how "time" in force = (change of momentum)/(time) is equal to 2x/u (the time it takes for the particle to travel to the opposite face and back again) -- I...

Thank you so much for your help jtbell -- this really cleared things up for me quite a bit! I really do appreciate it!

Jtbell, first of all, thank you for clarifying things up for me! But I'm not still quite sure what you mean by So in the case I mentioned in the original post, would we still state the measurement as 100.5mm instead of 101mm? But stating the uncertainty explicitly doesn't negate the fact that...

Firstly, thank you for your reply! I just wanted to ask -- wouldn't we have to round the 100.5mm to 101mm, since the uncertainty implies that the measurement is only accurate/precise to the nearest mm -- wouldn't giving the measurement to 1 decimal place create false precision/ suggest that the...

Hi, I've attached an image here which will hopefully clarify what I mean by how one end of the metal rod is at the 120mm graduation and the other is between 19mm and 20mm: In order to calculate the length of the metal rod, we have to approximate/estimate where the left end of the metal rod...