Correct Answer - A::C::D
(a.,c.,d.) `(KE + PE)_f = (KE + PE)_i` in all situations.
Hence, `KE_f` is also equal as `PE_f = 0`. Hence, all the particles collide with the same speed.
`-h = vt_1 -(1)/(2) "gt"_1^2` [for first particle] …(i)
`-h = -vt_2 - (1)/(2) "gt"_1^2` [for second particle] ...(ii)
From Eq. (i) and Eq. (ii), `t_2 gt t_1`
`t_2` = maximum, `t_1` = minimum
i.e., options ( c) and (d) are correct.