Question

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 v 1 . The same object is then raised again to the same height h but this time is thrown downward with velocity v1 v 1 . It now reaches the ground with a new velocity v2 v 2 . How is v2 v 2 related to v1 v 1 ?

Answer 1

Answer:

$\frac{v_{2}}{v_{1}} = \sqrt{2}$

Explanation:

Each object is modelled as follows:

First object

$v_{1} = \sqrt{2\cdot g\cdot h}$

Second object

$v_{2} = \sqrt{v_{1}^{2}+2\cdot g \cdot h}$

$v_{2} = \sqrt{4\cdot g \cdot h}$

The relation between $v_{2}$ and $v_{1}$ is:

$\frac{v_{2}}{v_{1}} = \sqrt{2}$