A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity `20 ms^-1`. The second stone will overtake the first after travelling a distance of `(g=10 ms^-2)`

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/2-gn)]^(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=1/2"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)