A woman stands near the edge of a cliff and drops a stone over the edge.

Question

A woman stands near the edge of a cliff and drops a stone over the edge. Exactly one second later she drops another stone. One second after that, how fast is the distance between the two stones changing? Assume that the position function of each stone is given by s(t) = −4.9t^2 m where t is in seconds.

Answer 1

9.8 meters per second

v(1)-v(2) = -9.8 + 19.9 = 9.8 m/sec

same answer as your 2 men on a cliff 10 m apart posted question