<p>Yeah...sorry megamike, I wasn't really thinking when I wrote that. The top of its trajectory is when v=o and since 0=vo-at, when a is greater, then t would be smaller, thereby making it take less time to rise and more time to fall. The particle must overcome more resistance to ascent when it is rising, but it falls much more freely during its decent.</p>
<p>For the mass, I got something E26 kg.</p>
<p>potpurri, going down does take longer. The test did ask for justification, and I certainly did not include proof by contradiction in my answer. However, although the answer pretty much has to be based on intuition, I was trying to make it clear in an extremely rigorous way - thus my rambling up above.</p>
<p>"The ball falls faster on the way down bc Fg works to accelerate in downward direction, wheras this decelerates motion in the upward direction."</p>
<p>Think about your reasoning for a second. Did you even use the fact that there is air resistance? If this was a valid justification, it would apply even if the air wasn't there, which can't be possible, since without resistance the time it takes the ball to rise and descend will be the same.</p>
<p>My equation was dv/dt = g - bv/m</p>
<p>I got that as well I think.</p>
<p>for e/m #2, what was the shape of the current graph? did it start off at 0 or did it approach 0?</p>
<p>I got the same DE as ohnoes. I'm pretty sure the current graph did not approach zero, though I'm also pretty sure I failed that problem, so I might not be the best person to believe there ;)</p>
<p>And the ball definitely took longer coming down.</p>
<p>i got that it started out as vo and approached zero like...inverse parabolic? and then when it hit zero at the 1/2 point it did the same thing again inverse parabolic (i mean the shape was like that not the actual equation) until Vt</p>
<p>oh but yea...i doubt i did it right</p>
<p>Rilla- that's exactly what I got. hopefully we both got it :) I'm pretty sure both the ends should be the same, but I didn't know what happened in the middle so I just randomly drew a parabolic shape.</p>
<p>...thanks guys! lol i got an inverse relationship but i had no idea what i was doing.</p>
<p>...I love you guys! You have made me feel a HELL of a lot better about my FRQ! Prior to this, I thought I totally BOMBED it!</p>
<p>Although, I still think it was damn hard lol.</p>
<p>Anyways, thanks!</p>
<p>Hey guys, for the diff. eq., does it matter if you isolated the derivative?</p>
<p>Also, when they asked for the diff. eq. is g negative? In the table of constants, g is 9.8 m/s^2 at the earth's surface so I put something to the effect of dv/dt=-g-kv/m </p>
<p>Thanks in advance for any insight</p>
<p>I'm pretty sure it can be negative. If the scoring guidelines for past FRQ's are any indication, it really doesn't matter whether or not you isolated the derivative to get creadit.</p>
<p>What did everyone get for the equipotential line on #1 and most of the questions on #2 for the E & M?</p>
<p>the equipotential line was perpendicular to the electric field and passed though the point they told you to make it pass through.</p>