# Two Stones

One stone is thrown upwards from a point on the ground with an initial velocity v1 and simultaneously another stone is dropped from a height. They reach the ground at the same instant. Compare the velocities which they have attained.

###### Solution.

The equation for velocity as a function of time is given by

vf=vi+a∙t

where vf – the final velocity of the object, vi – the initial velocity of the object, a – the acceleration and t – the time it takes the object to change its velocity.
We rewrite the equation for the first stone by renaming vi by v1 and by setting a=-g.

vf=v1-g∙t (1)

First let’s show that the first stone returns at the same magnitude of the velocity with which it left.
The velocity-position equation of motion of the object is given by

vf2=vi2+2∙a∙(xf-xi)

where xf – the final position of the object, xi – the initial position of the object, vf – the final velocity of the object, vi – the initial velocity of the object and a – the acceleration.

We rewrite the velocity-position equation of motion of the object by renaming xf by yf, xi by yi and by setting a=-g.

vf2=vi2-2∙g∙(yf-yi)

The initial position of the stone equals the final position of the stone yf=yi.

Hence,

vf2=vi2-2∙g∙0

or

vf2=vi2

From the last equation we can see that the final and initial velocities have equal magnitudes. However, we remember that the final and initial velocities have different directions.

Then,
vf=-vi

We can rename vi by v1 (first stone) and substitute it into the equation (1), we get

-v1=v1-g∙t

or

-2v1=-g∙t (2)

Now let’s look on the second stone. It was dropped from a height. The equation for velocity as a function of time is given by

vf=-g∙t

We rewrite the equation for the second stone by renaming vf by v2

v2=-g∙t (3)

Finally, we can compare equation (2) with equation (3), we take into account that the both stones spend the same amount of time in the air and we get

v2=-2v1