Question

12-6. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

Answer 1

Distance at t = 1 second: s implies acceleration = -g = -9.8 m/s^2 s = 1/2 at^2 s = 1/2 times 9.8 times 1^2 s = 4.9 meter at implies t = 2 second at s_A = 4.9 meter at t implies s_B = 1/2 times 9.8 times 2^2 s_B = 19.6 meter Distance b/w them implies s_B - s_A = 19.6 - 4.9 d = 14.7 meter Ans