Recent content by Lostone

No, more of a thinking it out for me kind of thing. The lab still wants the horizontal displacement. According to the graph provided on the website, when the air is turned "on", it uses a drag force of cv (v is the velocity and c is a constant). However, it does not give the constant...

Another one of those "thinking it out" things. Now the "Air" has been turned on. I'm thinking that the equation would change slightly (since it still wants the horizontal disposition) to: x-xo = vo cos(//)t + 1/2 at^2 Plug in the same set of variables (with the time changing due to the new...

...you mean all this time my -calculator- was set to the wrong thing? *headdeskthud* Okay, I reset it to degrees and got 251.02. Which is insanely close to the approximated 255 I was looking at on the graph. All in all, I'm way over-thinking this problem, aren't I?

Pardon me while I try and sound this out. In the simulation, the initial launch angle was 45 degrees. The initial launch velocity was 50m/s. This took 7.1 seconds to complete. x: final horizontal position xo: initial horizontal position vo: initial velocity t: time (\\): launch angle...

Learned to use them? They haven't even been mentioned in the book yet. I feel like a total fish out of water. The text has only so far barely even mentioned parabola. If I'm even remotely understanding this, the horizontal velocity will not change. So, if I assume that both the horizontal...

First off, I'm glad I stumbled upon this place. Secondly, taking physics as an online course is probably not the best judgment call I have made in recent times. However, I'm here now and I plan on sticking through with the course. My problems, however, are lying with the labs. Since they...