A ball is prouected vertical up from the top of a cliff of height `h` with a speed `v_(1)` Another ball is projected vertically up with a speed `v_(2)` from the bottom of the cliff, after a time `t_(0)` from the instantof projection of the first ball, When will the balls meet?.

Answer

`(2(h+v_(2)t_(0))+g t_(0)^(2))/(2{(v_(2)-v_(1))+g t_(0)})`

Explanation

Let the balls meer after a time `t'` measured form the instant of projection of the first ball. Since, the second ball is projected aftre a time `t_(0)` from the instant of projection of the first ball, the second ball falls freely for a time `t'-t_(0)`.
If the balls meet between the top and bottom of the cliff, for the first ball, `s=-s_(1), v_(0)=v_(1), A=-g` and `t=t'`
This yields `s_(1)=-v_(1)t' +(1)/(2_ g t'^(2)` (i)
For the second ball, `S=S_(2), v_(0)=v_(2), a=-g` and `t=(t'-t_(0))`
Hence `S_(2)=v_(2)(t'-t_(0))-(1)/(2) g (t'-t_(0))^(2)` (ii)
Referring tothe fig we have `s_(1)+s_(2)=h` (iii)
Substituting `s_(1)` from eq, (i). `s_(2)` from eq. (ii) in eq, (iii), we have`
`v_(1)t'+(1)/(2)g t'^(2) +v_(2)(t'-t_(0))-(1)/(2)-(1)/(2)g(t'-t_(0))^(2)=h`
This yields ` t'=(2(h+v_(2)t_(0))+g t_(0)^(2))/(2{(v_(2)-v_(1))+g t_(0)})`
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