A particle is projected vertically upwards and t second after another particle is projected upwards with the same initial velocity. Prove that the particles will meet after a lapse of `((t)/(2)+(u)/(g))` time from the instant of projection of the first particle.

Explanation

Let the particles meet `t_(0)` second after the projection of the first particle. Further suppose that they meet a height h from the ground.
For first particle. `h=ut_(0)-(1)/(2)"gt"_(0)^(2)`...(i) For second particle, `h=u(t_(0)-t)-(1)/(2)g(t_(0)-t)^(2)`....(ii)
From eqs. (i) `&` (ii) `ut_(0)-(1)/(2)"gt"_(0)^(2)=u(t_(0)-t)-(1)/(2)g(t_(0)-t)^(2)rArrut_(0)-(1)/(2)"gt"_(0)^(2)=ut_(0)-ut-(1)/(2)"gt"_(0)^(2)+"gt"_(0)t-(1)/(2)"gt"^(2)`
`rArr ut-"gt"_(0)t+(1)/(2)"gt"^(2)=0rArru-"gt"_(0)+(1)/(2)gt=0rArrt_(0)=((1)/(2)+(u)/(g))`