Recent content by mlbuxbaum

Thank you. If it wasnt for the help of you two guys, i would still be scratching my head. I have some polishing to do on the final copy, but the data is there now. And best of all, i actually learned a few things in the process...cool. Thanks again guys !

I have attached my file.

you are correct, that is what i needed to do. I just wanted to put the formula into excel that would calculate v0.

i entered in your formula and it doesn't equal what i got. What is 1404.37?

I used this in the column for my unkown Y value =y0+xtan\theta-.5g(x2/v02cos\theta) i used my calculated x values and the radian values for 15,30,45, and 60 degrees

thank you for your help. I inserted the equation into my x,y chart in the y column. I started my x column at 0 and y column at 3. I need 11 x,y points for my chart (per the instructor) so i divided 350 by 11, and each division workes out to be ~31.82. So now i have my x value to plug into the...

I worked the V0 for 15 degrees to be 157.9030425 30 degrees at 116.7051809 45 degrees at 107.5520743 and 60 degrees at 114.9329655 now i need to find out how to find the eleven (x,y) coordinants i need to use so i can plot a curve. I am inputting the formulas into my excel worksheet and...

i went to the tutor on campus yesterday and he walked me through the problem. The tutor had the same class, same instructor, same problem a few years ago. 12=3+(v0sin15)(350/v0cos15)-.5(32.2)(3502/v02cos215) so plugging numbers in and working it through 9=350tan15-16.1(131295.1043/v02)...

12=3+.2679x-(16.1x2/.9330v02 .9330(12-3-.2679x/-16x2)=v02 \sqrt{}.9330(12-3-.2679x/-16x2)=\sqrt{}v02 v0=\sqrt{}.9330(12-3-.2679x/-16x2)

i fixed the missing x. Finding v0 is where i am having my issue. I'm not sure how to get it alone

y=3+x(.2679)-16.1(x2/v02(.9330)) using 32.2 ft/sec2 for g

t=x/v0cos\theta0 i then substitute that value for t into the equation y=y0+xtan\theta0-.5g(x2/v02cos2\theta) ??

I don't know v0? Am i correct in assuming that the only values of x&y i have are (0,3) and (350, 12)? and that i plug those values into the equations to get the quadratic? Sorry if i am frustarting you. I am trying really. To find V0 would i divide both sides (the x equation) by t(cos\theta)...

I figured that .72 was sin46°(actually .7193), but in the eq he shows 4.9=vsin(46)/t 4.9=.72v/125/.7v not sure why the equation is divided by cos46°. my problem is alittle more meaty than this original.(to me) This is for my Intro to Engineering class, for the Engineering with Excel...