# two particles move in a uniform gravitational field with an acceleration g.at the intial moments particles are located at one point and moved with velocities v1=3 m/s and v2=4m/s horizontally in opposite directions.find the distance between the particles at the moments when their velocity vectors are mutually perpendicular.

Akash Kumar Dutta
98 Points
9 years ago

apply vector forms as i,j
and then put A.B=0
A=vel of part 1 at t time
B= ''    ''    ''    2  ''   '' ''

FITJEE
43 Points
9 years ago

2

SOURAV MISHRA
37 Points
8 years ago

take the point of projectionas the origin and the horizontal and the vertical directions as x and y axes.

let the initial velocities be 3i and -4i then the velocities at any time t are 3i - gtj and -4i - gtj respectively.

since these are perpendicular their scalar product is zero.

this gives 12 = g2t2 so that t = 0.35s

now we calculate the co ordinates of the particles.

these are given by 3ti - gt2/2j and -4ti - gt2/2j. so that the distance between them at any time t is given by 7t m.

therefore the required distance is 7*0.35 = 2.45m.

therefore the distance between them when their velocity vectors are perpendicular is 2.45m.