Two particles `1` and `2` are projected simultaneously with velocities `v_(1)` and `v_(2)`, respectively. Particle `1` is projected vertically up from the top of a cliff of height `h` and particle `2` is projected vertically up from the bottom of the cliff. If the bodies meet (a) above the top of the cliff, (b) between the top and bottom of the the cliff, and (c ) below the bottom of the cliff, find the time of meeting of the particles.
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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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