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. `5m`
B. `7sqrt(3)m`
C. `(7sqrt(3))/5m`
D. `7/2m`

Answer

c

Explanation

Let the particles moves perpendicular to each other at time t.

Hence `(4hati-"gt"hatj).(-3hatj-"gt"hatj)=0`
`implies -12+g^(2)t^(2)=0`
`implies t=sqrt((12/100))implies t=(sqrt(3))/5`
Hence distance,
`d=sqrt((ut+3t^(2))+((h-1/2"gt"^(2))-(h-1/2"gt"^(2)))^(2))`
`=7t=(7sqrt(3))/5 m`