4. A particle is launched vertically from the surface of the earth, and rises to a height of 2RE above the surface before falling back. Ignore air resistance entirely. a) Find the initial speed of the particle. (Look up the necessary numbers.) b) This is a problem of motion in one dimension, with a force that depends on position only. Find the time required to reach the turning point (maximum height). (RE is just the radius of the earth). This seeems so simple that I must be approaching this all wrong but here goes. a). Putting the origin at the earths surface, we get that mvo^2/2 = mgh, where h = 2RE. Therefore rearranginf for vo we get vo = 2sqrt(gRE). I'm skeptical whether this is right or not because this is a level 2 mechanics course, and it just seems to easy for the course and the prof. b), We known that U(r) - U(ro) = integral from ro to r of F dr, and thet U(ro) is zero, and U(r) = 2mgRE. F = mg, an then setting up an equation I get mdv/dt = 2mgRE. Does this seem right. I just feel that its no correct?