Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities u_(1)=3.0ms^(-1) and u_(2)=4.0ms^(-1) horizontally in opposite directions. Find the distance betwen the particles at the moment when their velocity vectors become mutually perpendicular.

Two particles move in a uniform gravitational field with an accelerati

1.	Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities u_(1)=3.0ms^(-1) and u_(2)=4.0ms^(-1) horizontally in opposite directions. Find the distance betwen the particles at the moment when their velocity vectors become mutually perpendicular.
Answer» Solution :The situation is shown fig. Let the velocity vectors become PERPENDICULAR after time t when both particles has fallen same vertical distance `1/2"GT"^(2)` and have acquired same vertical velocites gt. Let their RESULTANT velocities make angle `theta_(1)` and `theta_(2)` with horizontal. Then `tan theta_(1)=("gt")/(v_(1))` and `tan theta_(2)=("gt")/(v_(2))` Since velocity vectors are perpendicular `theta_(1)+theta_(2)=90^(@)`, hence `tan theta_(2)=cot theta_(1)` It makes `tan theta_(1)=("gt")/(v_(1))` and `cot theta_(1)=("gt")/(v_(2))` On multiplying we GET `t=(sqrt(v_(1)v_(2)))/g` and `D=(u_(1)+u_(2))t`