Answer to Question #173275 in Mechanics | Relativity for Jonie Matienzo

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 $v_i=v_0.$