Two bodies are thrown verically upward, with the same initially velocity of `98 m//s`, but the second one is thrown `4` sec. after first is thrown. If after `4n` seconds after the first particle is thrown, both of them meet then value of `n` is ?

Answer

`3`

Explanation

`98t - (1)/(2)"gt"^(2) = 98(t - 4) - (1)/(2)g(t - 4)^(2)`
Gives `t = 12 sec`.