Hi there again, I am having problems answering this question A girl wishes to estimate the depth d of a straight mine-shaft. She drops a stone vertically down the shaft and finds that there is an interval of 6 seconds between the instant she dropped the stone at the top of the shaft and the time that she heard the stone hit the bottom of the shaft. (i) She decides to take a first estimate of d by assuming that the stone took 6 seconds to drop from top to bottom of the shaft. Calculate this estimate. (ii) She then uses her first estimate together with the speed of sound, which is 332 m/s, to estimate the time taken for the stone to drop. Calculate this second estimate of d. (iii) Determine the actual depth of the shaft to the nearest metre. s=d u=0 v=/ a=9.8 t=6 using the following equation s=ut+1/2at^2 d=1/2 * 9.8 * 6^2 i get d=176.4m but iam not sure if this is correct though i am really stuck in this part of the question here are some random ideas i have tried t=d/s 176.4/332 = 0.53 s giving me the time taken for the sound to go up 6-0.53= 5.47 s time taken for the stone to drop s=d u=0 v=/ a=9.8 t=5.47s s=ut+1/2at^2 d=1/2 * 9.8 * 5.47^2 d=146.6m from here i dont know were to go with this question Any suggestions/ideas will be much appeciated. thanks in advance.