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.

Options

A. `(2sqrt(3))/(g)`
B. `(7sqrt(3))/(g)`
C. `(14sqrt(3))/(g)`
D. `(14sqrt(3))/(g)`

Answer

c

Explanation

`vecv_(A)=3t-gthatj`
`vecv_(B)=-4i-gt j`
When they are perpendicu lar
`vecv_(A)-vecv_(B)=0`
`therefore t=sqrt(12)/(g)=(2sqrt(3))/(g)`
`therefore` Distance between them d `=(v_(1x)+v_(2x))t`
`=(3+4)t=(14sqrt(3))/(g)`