A particle is projected vertically upwards from a point O with velocity V and a second particle is projected from a point P (at a height h above O) with velocity v and angle of projection `theta` . When is the distance between them is minimum.

### 1 Answer | Add Yours

A particle is projected vertically upwards from a point O with velocity V and a second particle is projected from a point P (at a height h above O) with velocity v and angle of projection theta. The time when the distance between them is minimum has to be determined.

At a time t after projection of the two, the following hold for the two particles. The height of the particle that was projected from O is equal to `v*t - (1/2)*g*t^2` . The height of the particle that was projected from P is equal to `h + v*sin theta*t - (1/2)*g*t^2` . The horizontal distance traveled by the particle from P is `v*cos theta*t` .

The distance between the two is:

D = sqrt((h + v*sin theta*t - (1/2)*g*t^2 - v*t + (1/2)*g*t^2)^2 + (v*cos theta*t)^2)

= `sqrt((h + v*sin theta*t - v*t)^2 + (v*cos theta*t)^2)`

This is minimum when D' = 0

D' = (1/2)*(1/sqrt((h + v*sin theta*t - v*t)^2 + (v*cos theta*t)^2))*(2*(h + v*sin theta*t - v*t)*(v*sin theta - v)+2*(v*cos theta*t)*(v*cos theta))

D' = 0

=> `2*(h + v*sin theta*t - v*t)*(v*sin theta - v)` +`2*(v*cos theta*t)*(v*cos theta)` = 0

=> `(h + t*(v*sin theta - v))*(v*sin theta - v)` + `v^2*cos^2 theta*t` = 0

=> `h*(v*sin theta - v)` + `t*(v*sin theta - v)^2` +`v^2*cos^2 theta*t` = 0

=> `t*((v*sin theta - v)^2+v^2*cos^2 theta)` = `h*(v - v*sin theta)`

=> `t*(v^2*sin^2 theta + v^2 - 2*v^2*sin theta+v^2*cos^2 theta)` = `h*(v - v*sin theta)`

=> `t*(v^2 + v^2 - 2*v^2*sin theta)` = `h*(v - v*sin theta)`

=> `t*(2v - 2*v*sin theta)` = `h*(1 - sin theta)`

=> `t*2v*(1 - sin theta)` = `h*(1 - sin theta)`

=> `t*2v = h`

=> `t = h/(2v)`

The distance between the particles is minimum after a time `t = h/(2*v)`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes