One stone is thrown upwards from a point on the ground with an initial velocity v_{1} 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

v_{f}=v_{i}+a∙t

where v_{f} – the final velocity of the object, v_{i} – 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 v_{i} by v_{1} and by setting a=-g.

v_{f}=v_{1}-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

v_{f}^{2}=v_{i}^{2}+2∙a∙(x_{f}-x_{i})

where x_{f} – the final position of the object, x_{i} – the initial position of the object, v_{f} – the final velocity of the object, v_{i} – the initial velocity of the object and a – the acceleration.

We rewrite the velocity-position equation of motion of the object by renaming x_{f} by y_{f}, x_{i} by y_{i} and by setting a=-g.

v_{f}^{2}=v_{i}^{2}-2∙g∙(y_{f}-y_{i})

The initial position of the stone equals the final position of the stone y_{f}=y_{i}.

Hence,

v_{f}^{2}=v_{i}^{2}-2∙g∙0

or

v_{f}^{2}=v_{i}^{2}

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,

v_{f}=-v_{i}

We can rename v_{i} by v_{1} (first stone) and substitute it into the equation (1), we get

-v_{1}=v_{1}-g∙t

or

-2v_{1}=-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

v_{f}=-g∙t

We rewrite the equation for the second stone by renaming v_{f} by v_{2}

v_{2}=-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

v_{2}=-2v_{1}