Saved Bookmarks
| 1. |
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.) |
|
Answer» Solution :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)` 
|
|