Two particles move in an uniform gravitational field with acceleration g. Initially the particles were located at one point and moved with velocities v1 = 3m/s and v2 = 4m/s horizontally in opposite directions. Find the distance between the particles when their velocities becomes mutually perpendicular.
283See the us assume the v1 is in positive and v2 is in negative direction. or vice versa.
g wil act down ward so it will change the in both particles.
after time t
Va= V1 - gt
Vb= -V2 - gt
now their velocity are perpendicular to each other.
dot product is zero
Va.Vb= 0
(V1 - gt).(-V2 - gt)=0
solving t=√V1V2/g
the distance between then will be Vrel*t
= {V1 - (-V2)}√V1V2/g
putting the values of v1 and v2 respectively
we get d = 14√3 / g
.
ARKADYUTI BANDYOPADHYAY Well,the answer is 2.5m,but since your answer is perfectly logical and the answer is NEARLY correct to the actual answer,I may as well consider YOUR solution and YOUR answer to be the correct one. Sahil Jain 14* 1.71 / 9.8 = 2.5 m
46614√3g
ARKADYUTI BANDYOPADHYAY Can you explain your answer?ARKADYUTI BANDYOPADHYAY I'm totally in the dark as to how you arrived at this equation.Himanshu Giria is it right ??
187You have to use the concept that dot product of two perpendicular vectors is 0. assume that they will become perpendicular after time 't'. treat the point from where they are are projected be the origin. so,
(-v1i -gtj).(v2i - gtj) = 0
solve for t and then find the distance, you will get the answer.