Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities `v_1 = 3m//s and v_2= 4m//s` horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Answer

`l=(v_1+v_2)sqrt(v_1v_2//g)=2.5m`

Explanation

Let the velocities of the particles (say `vec(v_1)` and `vec(v_2)`) becomes mutually perpendicular after time t. Then their velocities become
`vec(v_1)=vec(v_1)+vec(g t) , vec(v_2)=vec(v_2)+vec(g t)` (1)
As `vec(v_1)_|_vec(v_2)` so, `vec(v_1)*vec(v_2)=0`
or, `(vec(v_1)+vec(g t))*(vec(v_2)+vec(g t))=0`
or `-v_1v_2+g^2t^2=0`
Hence, `t=(sqrt(v_1v_2))/(g)`
Now from the Eqs. `vec(r_(12))=vec(r_0)_((12))t+1/2vecw_(12)t^2`
`|vec(r_(12))|=|vec(v_0)_((12))|t`, (because here `vecw_(12)=0` and `vec(v_0)_((12))=0`)
Hence the sought distance
`|vec(r_(12))|=(v_1+v_2)/(g)sqrt(v_1v_2)` (as `|vec(v_0)_((12))|=v_1+v_2`)