## Essential University Physics: Volume 1 (3rd Edition)

a) We first must find the final y-velocity: $v_f=\sqrt{v_0^2 +2a_y\Delta y}$ $v_f=\sqrt{17.7^2 -2(9.81)(6.42)}=13.7\ m/s$ We use conservation of momentum to find the velocity after they stick: $v = \frac{13.7\times .24}{.114+.24}=9.3\ m/s$ We now find the overall height: $h_{max}=7.65 + \frac{(9.3)^2}{2(9.81)}=12.1 \ m$ b) They start at 12.1 meters high and fall, so we find: $v_f=\sqrt{2(9.81)(12.1)}=15.4\ m/s$