Two men stand 10 m apart on level ground near the edge of a cliff.

Question

Two men stand 10 m apart on level ground near the edge of a cliff. One man drops a stone and one second later the other man drops a 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

change in distance is velocity

take the derivative of the distance function

v(t) =s'(t) =-9.8t

v(1) = -9.8 m/s

v(2) = -19.6 m/s

v(1) - v(2) =9.8 m/s = change in distance of the two stones after that 2nd second

(assuming both stones have not hit the ground yet. -4.9(2^2) = -4.9(4) = -19.6

19.6 meters below is where the 1st stone is after 2 seconds. the cliff has to be higher than 19.6 meters )