All the particles thrown with same initial velocity would strike the g

1.	All the particles thrown with same initial velocity would strike the ground. .
Answer» with same speed. simultaneously time would be least for the particle thrown with velocity `v` DOWNWARD i.e., particle `1`. time would be maximum for the particle `2`. Solution :(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.

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