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 byA. `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/2-gn)]^(2)`D. `g/5[(u-gn)/(u/2-gn)]^(2)`

A stone is dropped from the top of a tall cliff and `n` seconds later

1.	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 byA. `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/2-gn)]^(2)`D. `g/5[(u-gn)/(u/2-gn)]^(2)`
Answer» Correct Answer - A Let the two stones meet at time t. For the first stone, `S=1/2"gt"^(2) (therefore u=0)`……………(i) For the second stone, `s_(2)=u(t-n)+1/2g(t-n)^(2)`…………….(ii) Displacement is same `therefore S_(1)=S_(2)` `1/2"gt"^(2)=u(t-n)+1/2(g(tn)^(2)` (Using (i) and (ii)) `1/2"gt"^(2)=ut-nu+1/2"gt"^(2)+1/2"gn"^(2)-"gtn"` `ut-"gtn"="nu"-1/2"gn"^(2)` `t=("nu"-1.2"gn"^(2))/(u-"gn") = (n(u-g)/(2n))/(u-gn)` or `S_(1) = g/2[(n(u-(gn)/(2)))/(u-gn)]^(2)` from (i)

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