Recent content by agm1984

anyone? which bridge looks more legit?

Homework Statement i am building a popsicle stick bridge, and i basically have two trusses, which i am attaching together to make the bridge (each are one side of the bridge), and I am having trouble deciding which way to orient the two trusses. if you look at the picture below, you can...

Try "coincidences of math" google search maybe. Physics and hard science are typically oriented around objective quantifiable aspects and not so much mystical connections. Math and geometry probably have some coincidences, and probably some aspects of calculus are coincidental. As the poster...

hmm i seem to have done an arithmetic error because that does net the correct answer. d= vot + 0.5at2 d=(25.981)(0.8) + 0.5(-9.8)(0.8)2 d=17.6488~ and we can round that up with some sig figs (my book is all over the place but clearly they are taking 3 sig figs for 17.7m) sorry if...

i don't know, i am very confused and frustrated at the moment. the max height of the ball is vf2 = vo2 + 2ad vf = 0 vo = 30sin60 or 30cos30 = 25.981m/s a=-9.8m/s d=unknown this is what we are solving for t=unknown vf2 = vo2 + 2ad d=34.44m after this, we want to see how high the...

that equation is incorrect because a=-9.8 is not true in the horizontal motion.

the pebble is slingshotted 30 degrees to the vertical at 30meters/sec so the resultant force is 30m/s aimed 30 deg to the vertical (aka 60 degrees to the horizontal, aka vertical force = 30sin60=25.981m/s and horizontal force = 30cos60 = 15m/s or if you draw the triangle the other way the Y...

i don't see how that would be incorrect. it is for the horizontal component. the projectile is moving horizontally 12 meters with velocity 15m/s because the actual force is 30m/s 30deg to vertical, thus horizontal force is (30cos60) = 15m/s therefore it will take 0.8 seconds to go 12meters

thats right, i was thinking illogically with that attempt. i now believe the horizontal component is simply d=v*t which solves t for 0.8seconds to reach 12m i don't know what to do with the vertical distance though. i am frustrated now.

im gaining confidence that the time to go 12m is 0.8 horizontal: d=vt 12m=15m/s*t t=0.8 so i just need to get the right vertical distance... a=-9.8m/s d=? t=0.8 v original=30sin60=25.981~ v final=? a,t,vo=d so it looks like i really need to use d=vo(t) +1/2at2 but why does that give 23.9208m...

i think I am spinning my wheels here

lol ok nope that failed as well, and i got vertical d=23.92m by that method.

hmm i got another answer using d=vot +1/2at2 again but for vertical, and i got d=14.07m still off im going to try switching the equations around. d=vt for horizontal and d=vt+1/2at^2 for vertical.

hmm this seems to have failed but also seems close. i did d=vot +1/2at2 and solved for t, t=[2(d-vo)]/a t=2(12m-15m/s)/-9.8m/s2 t=0.612seconds then i attempted to get the vertical distance after, using vertical velocity (30sin60): d=vot d=[(30)(sin60)]*(0.612) and got d=15.91 approx the...

ok here's my plan of attack I am about to try. im going to see if i can calculate how long it takes the projectile to travel 12 meters horizontal, and then I am going to see if i can figure out how high it will be after that amount of time.