# Relative velocity

1. Feb 23, 2013

### Litcyb

1. The problem statement, all variables and given/known data

Ball 1 is dropped from rest at a height of 10m above the ground. at the same time, ball 2 is thrown straight upward from ground level with an initial speed of 10m/s.

2. Relevant equations

find the relative velocity of the balls when they pass each other at t=1 second?

equation given, relative velocity ==> v12= lv2-v1l

vf=vi+at

3. The attempt at a solution

since ball1 is falling, in respect to earth, the acceleration is possitive thus, v1f=v1i+at equals, vf= 0+9.8m/s^2 *1s

vf= 9.8 m/s

for ball 2 the ball is thrown upwards, thus being in a negative acceleration in respect to earth thus,

v2F=v2i+a*t
V2f= 10m/s+ (-9.8m/s^2)(1s)

v2f= 0.2

now, lV2-V1l = 9.6m/s

now, the correct answer is 10m/s

and i wonder how come and why??

could it be that in respect to each other, their acceleration is negative? thus, resulting in

v1=-9.8
v2=0.2

lv2-v1l = l0.2+9.8l = 10m/s?

2. Feb 23, 2013

### G01

Remember that the balls are traveling in opposite directions so the velocities have opposite signs.

3. Feb 23, 2013

### Litcyb

so if i initially calculated that v1= 9.8 and v2= 0.2 their opposite signs would be -9.8 and -0.2 and when applied to the equation lv2-v1l = l -0.2+ 9.8l is still 9.6 m/s :-/

and for some reason the professor uses, v(f) = vi-gt

why is he subtracting gravitational force * time???????? shouldnt it be adding

Last edited: Feb 23, 2013
4. Feb 23, 2013

### G01

You misinterpreted my above statement. If one ball is moving down and the other ball moves up then, v_up will have the opposite sign of v_down. Which one is positive and which is negative will depend on your choice of coordinate system.

Your professor is probably using a convention where g is always positive. So, he changes the sign in the equations when necessary instead of the sign of g itself.

Also, note that g is the acceleration due to gravity, not the force of gravity.