Relative Velocity of Balls: Solving for t=1 sec

[Litcyb]

Homework Statement

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.

Homework 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

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?

 

[G01]

Homework Statement

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.

Homework 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

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?

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

 

[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? shouldn't it be adding

 

[G01]

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? shouldn't it be adding

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.

 

[Nickie]

I would first clarify the question with the person who gave the homework. It seems that there may be some confusion about the direction of the velocities and the reference frame being used. If we assume that the velocities are with respect to the ground, then the relative velocity at t=1 second should be 9.6 m/s. However, if we assume that the velocities are with respect to each other, then the relative velocity should be 10 m/s. It is important to clearly define the reference frame and direction of velocities in order to accurately solve this problem.