An Olympic skier makes a run down a snowy hill starting at 100 m vertical distance above the ground. (a) If friction is neglected, what is the velocity of the skier at the bottom of the hill? (b) What is the velocity half way down the hill?

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An Olympic skier makes a run down a snowy hill starting at 100 m vertical distance above the ground. (a) If friction is neglected, what is the velocity of the skier at the bottom of the hill? (b) What is the velocity half way down the hill?

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Answer 1 · 2017-12-19T13:05:17

$(a)$ $(a)$ The velocity at the bottom of the hill is $= 44.3 m s^{- 1}$ $=44.3ms^-1$ . $(b)$ $(b)$ The velocity half way is $= 31.3 m s^{- 1}$ $=31.3ms^-1$

Explanation:

The potential energy of theskier at the top of the hill is

$P E = m g h$ $PE=mgh$

The kinetic energy at the bottom of the hill is

$K E = \frac{1}{2} m v^{2}$ $KE=1/2mv^2$

As,

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

$v^{2} = 2 g h$ $v^2=2gh$

$v = \sqrt{2 g h}$ $v=sqrt(2gh)$

Plugging in the values of $h = 100 m$ $h=100m$ and the acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The svelocity of the skier at the bottom of the hill is

$v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 100} = 44.3 m s^{- 1}$ $v=sqrt(2gh)=sqrt(2*9.8*100)=44.3ms^-1$

To calculate the velocity half way the hill plug in $h = 50 m$ $h=50m$

$v = \sqrt{2 g \cdot 50} = 31.3 m s^{- 1}$ $v=sqrt(2g*50)=31.3ms^-1$

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An Olympic skier makes a run down a snowy hill starting at 100 m vertical distance above the ground. (a) If friction is neglected, what is the velocity of the skier at the bottom of the hill? (b) What is the velocity half way down the hill?

1 Answer

Explanation:

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