Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;

where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
y = 10.44cos(2t - 0.291) cm
Explanation:
y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)
v = y' = -2Αsin(2t + φ)
10 = Acos(2(0) + φ) = Acosφ
6 = -2Αsin(2(0) + φ) = -2Asinφ
6/10 = -2Asinφ/Acosφ = -2tanφ
tanφ = -0.3
φ = -0.291 radians
10 = Acos(-0.291)
A = 10/cos(-0.291) = 10.44
"Free fall" means that gravity is the ONLY force acting on the object. If there's air resistance, then it isn't free-fall.
Objects that fall near earths surface are ALWAYS falling through air, unless they're inside some kind of a vacuum chamber with all the air removed from it.