Recent content by Brentavo7

I am not saying that G is in error. At least not on a planetary level. I am saying that those slight differences at our level could show that there is a variable somewhere that may effect G greatly at smaller scales.

How can coulombs law have nothing to do with mass I thought force equals mass times acceleration. It would seem to play some role. Masses all contain charged particles so I figured there should be some + and - interactions on an atomic level. They just cancel each other on a large scale...

Thank you very much for replying. It may sound silly but could a bounce not be considered a temporary repulsion. If gravity somehow aligns the atoms similar to an electromagnet then the part touching the ground would always be the opposite polarity of the ground. But when a bounce occurs the...

Please don't be too cruel. I obviously have to be doing something wrong but it seems to me that the gravitational constant may be a variable based on coulombs constant and an objects charge per mass. The formula I have found would be G = Ke x ((Q1 / M1) x (Q2 / M2)) This would give a...

(Vox^2 + Voy^2) - (14.7t*Voy) + (48.02t^2) = 0 2400.9998 - 698.904t + 48.02t^2 = 0

yeah I am not too fond of them either. I took this equation (Vox*t)(Vox) + (Voy*t-.5*g*t^2)(Voy-g*t) = 0 divide both sides by t you get Vox^2 + (Voy - .5*g*t)(Voy - g*t) = 0 you can create a quadratic with this Vox^2 + Voy^2 - (.5*g*t*Voy) - (g*t*Voy) + (.5*g^2*t^2) plug in g= 9.8; Vox=...

the projectile is 128.87m away from the launch point at the half way point. But 130.705m away at the max.

use the formula I had in post 21 for question c. your Vox and Voy are correct. I used this site to cheat. http://www.csgnetwork.com/quadraticcomplex.html. when you get the time just figure out your x and y at that time. AJ had it earlier on post 22. then use x^2 + y^2 = d^2 to get your distance...

Looks like it was mine. glad I got the other answers right with it. Thanks for all the help I appreciate it.

Good catch on the rounding. Thanks. now if I can just figure out my mistake on the angle I am all set.

really? what am I doing wrong. it seems simple enough. ArcSin ((square root of 2 * 2)/3) = arcsin ((1.414214 * 2)/3) = arcsine (2.828427/3) =Arcsine 0.942809 = 70.31.44 degrees according to my calculator.

Here are the answers I have a. range of projectile: 115.020m b. total time of flight: 9.70295s c. time in which the projectile is farthest from its launch point: 5.55736s d. farthest distance the projectile is from its launch point: 130.705m e. the critical angle theta c such that for...

Vox^2 + (Voy2 - 14.7*Voy + 48.02) Im getting confused with the sine and cosine of the angles

nope still confused on the angle.

with t= 4.85147 the distance from the launch point is 128.87m using x^2 + y^2 = d^2 at t= 5.55736 the distance from the launch point is 130.705m