a stone is dropped from the top of a tall cliff and 1 sec later a second stone is thrown vertically down with a velocity of 60 ft/sec. how far below the top of the cliff will the second overtake the first?

Question

mrnood · Answer

the distance fallen s= ut +0.5 at^2 in both cases

In the first case u = 0
in the second case  u = 60

When they cross the  distance s is the same -so you can solve for t

Then put that value of t in the first equation, and solve for s