A particle `P` is to be projected from a fixed point `A` on the ground with a fixed speed `u`. Another particle `Q` is also to be projected from a point `B` which is directly above the point `A`. The trajectory of the particle `Q` touches all possible trajectories (for different angle of projection) of the particle `P` in the vertical `x-y` plane. Find the equation of trajectory of the particle `Q`. (Take the point `A` as origin, horizontal axis `x-` axis and vertical axis `y-`axis Consider the figure.)

Explanation

In order for the trahectory of `Q` to touch that of `P` when the latter is projected vertically upward, `Q` must be projected horizontally from the highest point. To visualize this, consider the trajectories of `P` as the angle of projection of `P`(`theta`,say), approaches `90^(@)` (i.e. for vertical projection).Let `Q` be projected horizontally from `B` with a speed `v`, where `A=(u^(3))/(2g)`. The point of impact of `Q` on the horizontal plane represents the point of maximum range for `P` along the horizontal.
`OC=(u^(2))/(g)=vsqrt((2h)/(g))`, where `h=(u^(2))/(2g)`
i.e., `v=u`
The trajectory of `Q` is
`(u^(2))/(2g)-y=(1)/(2)"gt"^(2)`. where `t=x//v(v-=u)`