A stone is dropped from the top of a tall cliff and `n` seconds later another stone is thrown vertically downwards with a velocity `u`. Then the second stone overtakes the first, below the top of the cliff at a distance given by

Options

A. `g/2[(n(u-(gn)/2))/((u-gn))]^2`
B. `g/2[(n(u/2-gn))/((u-gn))]^2`
C. `g/2[(n(u/2-gn))/((u-(gn)/2))]^2`
D. `g/5[((u-gn))/((u/2-gn))]^2`

Answer

A

Explanation

Let the two stones meet at time t .
For the first stone ,
`S_(1)=1/2" "( :.u=0)" "...(i)`
For the second stone,
`S_(2)= u(t-u)+1/2g(t-n)^(2)`
Displacement is same ,
`:. S_(1) = S_(2) ,1/2"gt"^(2) = u(t-n)+1/2g(t-n)^(2)`
`1/2 "gt"^(2) ="ut" -"nu "+1/2 "gt"^(2) +1/2"gn"^(2) -"gtn" ,"ut" -"gn"="nu" -1/2"gn"^(2)`
`t = ("nu"-1/2"gn"^(2))/(u-"gn")=(n(u-g/2n))/(u-"gn")`
Substituting value of t in (i) , we get,
`S_(1) =1/2xxg[(n(u-("gn")/2))/(u-"gn")]^(2)`