Height of tap = 5m and (g) = \\\\(10 m/sec^{2}\\\\)For the first drop, \\\\(5=ut+gt^{2}\\\\)\\\\(=(0\ imes t)+\ imes 10t^{2v}=5t^{2}\\\\) or \\\\(t^{2}=1\\\\) or t=1.It mean that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.Distance covered by the second drop in 0.5 sec.\\\\(=(0.5)^{2}=1.25m\\\\)Therefore, distance of the second drop above the ground\\\\(=5-1.25=3.75 m\\\\)