# Can someone answer this physics question?

## A brick with mass of $m$ is released from the fourth floor of a building, and the brick hits the ground with a velocity of $v$. Look away from air restistance. From which floor would you have let go of the block if one wanted to hit the ground with double speed? (a) sixth floor, (b) 8th floor, (c) tenth floor, (d) twelfth floor or (e) sixth floor.

May 10, 2018

$16$th floor

#### Explanation:

The idea of free fall is that all gravitational potential energy will be converted into kinetic energy during the fall. When the brick strikes the ground, there is no potential energy remaining as the height of the brick is zero.

Mathematically, gravitational potential energy is given as

$\implies {E}_{p} = m g h$

Kinetic energy is given as

$\implies {E}_{k} = \frac{1}{2} m {v}^{2}$

Setting them equal, we get

${E}_{p} = {E}_{k}$

$m g h = \frac{1}{2} m {v}^{2}$

$g h = \frac{1}{2} {v}^{2}$

$\implies {h}_{v} = \frac{1}{2 g} {v}^{2}$

This result shows us the relationship between the initial height of the brick and the speed it will be traveling if all potential energy is converted into kinetic energy.

So if we want the brick to hit the ground with twice the speed $2 v$, then

$\implies {h}_{2 v} = \frac{1}{2 g} {\left(2 v\right)}^{2} = \frac{2}{g} {v}^{2}$

Comparing the heights:

${h}_{2 v} / {h}_{v} = \frac{\frac{2}{g} {v}^{2}}{\frac{1}{2 g} {v}^{2}}$

${h}_{2 v} / {h}_{v} = \frac{2}{\frac{1}{2}}$

${h}_{2 v} / {h}_{v} = 4$

$\implies {h}_{2 v} = 4 {h}_{v}$

So for the brick to hit the ground moving twice as fast, we need to quadruple the initial height! So if we originally dropped the brick at the $4$th floor, now we need to drop it at the $16$th floor.