Question

 Show that the velocity of impact at the ground of an object thrown

vertically upward is equal to the initial velocity of the upward throw.

Answer 1

From the law of conservation of energy the total energy of an object remains constant.

At the initial moment the total energy is equal to the kinetic energy:

$E_k=\frac{mv_0^2}{2},$

where m - the mass of object;

$v_0$ - the initial velocity of the upward throw.

At the highest point of trajectory the kinetic energy is zero and the total energy is equal to the potential energy:

$E_p=mgh,$

where h - the altitude of object.

Then the object begins to drop downward and on graond level its total enrrgy is equal to he kinetic energy again:

$E_k=\frac{mv_i^2}{2},$

where $v_i$ - the velocity of impact at the ground.

So, $\frac{mv_i^2}{2}=\frac{mv_0^2}{2},$

and $v_i=v_0.$