76. (a) For the 0.50 meter drop in “free-fall”, Eq. 2-16 yields a speed of 3.13 m/s. Using this as the “initial speed” for the final motion (over 0.02 meter) during which his motion slows at rate “a”, we find the magnitude of his average acceleration from when his feet first touch the patio until the moment his body stops moving is a = 245 m/s2. (b) We apply Newton’s second law: Fstop – mg = ma Fstop = 20.4 kN.
