Question

A ball is dropped from a building and simultaneously another ball is projected upward with the some velocity. Describe the change in relative velocities of the balls as a function of time.

Answer 1

Hii dear,

# Answer- Relative velocity = constant

# Explaination-
Let first ball be dropped from edge of the building and second ball be thrown vertically upwards with velocity u.

At t=0, velocity of 1st ball w.r.t. 2nd ball is
u12 = 0-(-u)
u12 = u

At t=0, acceleration of 1st ball w.r.t. 2nd ball is
a12 = g-g
a12 = 0

Relative velocity of 1st ball wrt 2nd ball after final time interval t is
v12 = u12 + a12.t
v12 = u + (0)t
v12 = u ...constant

Hence relative velocity of balls wil be always constant during their motion which is equal to velocity of projection of 2nd ball.

Hope this helped...

Answer 2

For first ball u= u1m/s

V=V1 m/s

a=g m/s 2

t=t sec

On substituting  the above values in equation given below we get :

V=u+at

V1=u1+gt1--------------------(1)

For second ball :

U= u2 m/s

V=v2 m/s

A=-g m/s2

T= t2 sec

Substituting all the above values in the following equation we get :

V=u+at

V2=u2-gt2---------------(2)

on subtracting Equation 1 – equation 2

--> we get

V1- v2= (u1-u2)+gt1+gt2

(V1-v2)-(u1-u2)= g(t1+t2)

Since u1=0

(V1-v2)-(0-u2)= g(t1+t2)

∴The change in final relative velocities and initial relative velocities  of two ball is =function of time