# Question #b6d23

May 16, 2016

25%

#### Explanation:

If we define the initial velocity as having magnitude $u$, then the vertical component of this is $u \cdot \sin 30$ (the horizontal component would be $u \cdot \cos 30$ but we are not interested in this as it has no effect on potential energy)

A word of caution: there is no definition or reference in the question as to what 30 degrees is measured from; it could be from the horizontal or the vertical. | have assumed from the horizontal. Please let me know if this is not the case and I will re-work.

The body will travel a maximum distance $s$ upwards. At this point its vertical velocity will be nil. We can used the following to determine $s$:
${v}^{2} = {u}^{2} + 2 a s$
We know $v$, final velocity is nil, and $u$, initial velocity is $u \sin 30$ and $a$ will be acceleration due to gravity, $g$ (which will be negative relative to the direction of the initial velocity). Hence:
$0 = {u}^{2} {\sin}^{2} 30 - 2 g s$
$2 g s = {u}^{2} {\sin}^{2} 30$
$s = \frac{{u}^{2} {\sin}^{2} 30}{2 g}$

So final potential energy will be:
$P E = m g h = m g s$

$P E = \left(m g\right) \cdot \frac{{u}^{2} {\sin}^{2} 30}{2 g}$

$P E = \frac{m \cdot {u}^{2} {\sin}^{2} 30}{2}$

Initial kinetic energy is
$K E = \frac{1}{2} m \cdot {u}^{2}$
So percentage conversion is:

$\text{PE"/"KE} = \frac{m \cdot {u}^{2} {\sin}^{2} 30}{2} / \left(\frac{1}{2} m \cdot {u}^{2}\right)$

$\text{PE"/"KE} = {\sin}^{2} 30 = 0.25$