# Question #3c3f8

Dec 18, 2016

The derivation is performed below...
The desired result for the velocity is $v = \sqrt{2 g h}$

#### Explanation:

In applying conservation of energy we will start from the configuration in which the arrow is drawn back and stationary in the bow.

Its potential energy (PE) at this point is $\frac{1}{2} k {d}^{2}$, using the symbols provided in the question.

Even though it gains considerable speed upon release, before finally slowing again to a halt we can "skip over" all of this because the problem does not concern itself with this part of the motion. This is one of the beauties of using conservation of energy in solving problems in mechanics.

When it comes to rest at height h, its energy is all gravitational PE, given by $m g \left(h + d\right)$. Note we add the extra term in "d" due to the fact that total upward travel of the arrow is h+d and not just h

Cons of energy insists that $\frac{1}{2} k {d}^{2}$ = $m g \left(h + d\right)$

Solve for h: $h + d = \frac{k {d}^{2}}{2 m g}$

And finally $h = \frac{k {d}^{2}}{2 m g} - d$

For the second result, we note that the arrow drops only a distance h in returning to "head level" (as opposed to h+d). The PE decrease is $m g h$ and the kinetic energy increase, equal in magnitude, is given by $\frac{1}{2} m {v}^{2}$

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

Cancel m, and solve for v:

$v = \sqrt{2 g h}$