Recent content by N8G

That being said, I figured out how to separate the variables but the integrals for part 2 and 3 both turned out to be horrendous given that I’m looking for the velocity wrt height functions. Each integral needed either aggressive attempts at u substitution or partial fraction decomposition...

Never mind. I think I lost my mind and forgot about how division works. I should be able to just say: mg - cv = mv dv/dy 1 = mv/(mg-cv) dv/dy dy = mv/(mg-cv) dv Sorry for that.

My bad, any x's should be y's. For the second part of the problem I have: mvdv/dy = mg - cv which I reduce to dv/dy = g/v - c/m From here I don't see a way to isolate the v term on the rhs from the dy when separating my variables. And I am taking g to be -9.8m/s^2

Homework Statement A professional thrower projects a football straight up in the air. 1. Assuming there is no air drag on the football, find the speed of the football as a function of height as the ball goes up. 2. Assuming the air drag on the football varies linearly with speed, find the speed...

Homework Statement Griffiths Introduction to Electrodynamics 4th Edition Example 1.10 Check the divergence theorem using the function: v = y^2 (i) + (2xy + z^2) (j) + (2yz) (k) and a unit cube at the origin. Homework Equations (closed)∫v⋅da = ∫∇⋅vdV The flux of vector v at the boundary of the...

(v^2)/(2g) = 400/(2*9.81) = approximately 20 Okay awesome. I must've made an egregious calculator error in the beginning. Thank you!

Yeah, I know that vf = vo +at however that gives me 0 = 20 + (-9.81)t and solving for t gives me a time of 2.039 seconds. However the work that I had already done with energy conservation told me that the max height of the object, that coincides with the point at which its velocity becomes zero...

Homework Statement Object launched vertically with v initial =20 m/s. No air resistance. How long does it take to reach its max height? Homework Equations 1/2 at^2 + volt = dy .5mv^2=mgh The Attempt at a Solution I began by using conservation of energy to find the max height. .5mv^2 = mgh...