Question

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?




help mee..

Answer 1

s = distance from top of the cliff
both the stones will cover same distance 's' until 2nd stone overtakes 1st.

since 2nd stone was released 1 sec. later,
let the time taken by 1st stone be 't' and time taken by 2nd stone be 't-1'

initial velocity 'u' for 1st stone = 0
and for 2nd stone, u =60 ft/sec
g= 32.174 ft/sec^2