A clock is thrown straight upward on an airless planet with uniform gravity \(g\), and it falls back to the surface at a time \(t_{f}\) after it was thrown, according to clocks at rest on the ground.

(a) Using the clock's motion as derived in Section 3.7, how much more time than \(t_{f}\) will have elapsed according to this moving clock, in terms of \(g, t_{f}\), and \(c\), the speed of light?

(b) Now suppose that instead of the freely-falling motion used in part (a), the moving clock has constant speed \(v_{0}\) straight up for time \(t_{f} / 2\) according to ground clocks, and then moves straight down again at the same constant speed \(v_{0}\) for another time interval \(t_{f} / 2\), according to ground clocks. How much more time than \(t_{f}\) will have elapsed according to this moving clock, in terms of \(v_{0}, g, c\), and \(t_{f}\) ?

(c) Now find the value of \(v_{0}\), keeping \(g\) and \(t_{f}\) fixed, which maximizes the final reading of the moving clock described in part (b). Then evaluate the final reading of this moving clock in terms of \(g, t_{f}\), and \(c\), and show that it is less than the final reading of the freely-falling clock described in part (a). (This is a particular illustration of the fact that the path which maximizes the proper time is that of a freely-falling clock, i.e., a clock that moves according to Newton's laws. The reader could choose some alternative motion for a clock, and show again that as long as it returns to the beginning point at \(t_{f}\) according to ground clocks, its time will be less than that of the freely-falling clock of part (a).)