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

B

Explanation

`V_1 * V_2 = 0` when` v_1 _|_ v_2 `
`:. (u_1 + a_1t) * (u_2 + a_2t) = 0`
`:. (3hati - 10thatj) * (4hati - 10t hatj) = 0`
or `t = (sqrt(0.12)) s`
Now in vertical direction they have no relative
motion and in horizontal direction their velocities
are opposite.
`:. d = 3t + 4t = 7t`
`=(7) (sqrt(0.12)) m`
`~~ 2.5m`