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    				| 1. | 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» Correct Answer - B `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` | |