Two particles are simultaneously released from points `A and D` as shown is Fig.4.41. How shold the value of (H) be adjusted inorder that the two particles collide?
Neglect sissipative forces.
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Explanation

If both particles collide, the time of travel for both particles should be equal.
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Path A-B : `T_(1)=t_(AB)+t_(BC)`
For Path AB : `S_(AB)=t_(AB)^(2) rArr (h)/(sin theta) =(1)/(2)g sin theta.t_(AB)^(2)`.
`t_(AB)=(1)/(sin theta)sqrt(2h)/(g)`
Velocity of the particle when it reaches `B`,
`v_(B)=0+(g sin theta).(1)/(sin theta) sqrt((2h)/(g))`
Path `AB`: As acceleration of the particles is zero, time to cover destance`BC`,
`t_(BC)=(BC)/(v_(B))=(h)/(sqrt2gh)=sqrt((h)/(2g))`
Hence, `T_(1)=(1)/(sin theta) sqrt(2h)/(g)+sqrt((h)/(2g))`
Path `DC`: The particle will free fall,
`H=(1)/(2)g t_(2)^(2) rArr T_(2) =aqrt(2H)/(g)`
For particle collide `T_(1)-T_(1)`, which gives
`H=(h)/(sin^(2) theta)(2+sin theta)^(2)`.