Question #b6d23

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Question #b6d23

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Answer 1 · 2016-05-16T18:42:26

25%

Explanation:

If we define the initial velocity as having magnitude $u$ $u$ , then the vertical component of this is $u \cdot sin 30$ $u*sin30$ (the horizontal component would be $u \cdot cos 30$ $u*cos30$ 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 $s$ $s$ upwards. At this point its vertical velocity will be nil. We can used the following to determine $s$ $s$ :
$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$
We know $v$ $v$ , final velocity is nil, and $u$ $u$ , initial velocity is $u sin 30$ $usin30$ and $a$ $a$ will be acceleration due to gravity, $g$ $g$ (which will be negative relative to the direction of the initial velocity). Hence:
$0 = u^{2} {sin}^{2} 30 - 2 g s$ $0=u^2sin^2 30-2gs$
$2 g s = u^{2} {sin}^{2} 30$ $2gs=u^2sin^2 30$
$s = \frac{u^{2} {sin}^{2} 30}{2 g}$ $s=(u^2sin^2 30)/(2g)$

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

$P E = (m g) \cdot \frac{u^{2} {sin}^{2} 30}{2 g}$ $PE=(mg)*(u^2sin^2 30)/(2g)$

$P E = \frac{m \cdot u^{2} {sin}^{2} 30}{2}$ $PE=(m*u^2sin^2 30)/(2)$

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

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

$\frac{PE}{KE} = {sin}^{2} 30 = 0.25$ $"PE"/"KE"=sin^2 30=0.25$
So the answer is 25%

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Question #b6d23

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