## Related questions with answers

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a $30^{\circ}$ slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.

Solution

VerifiedLet's first find the general formula for time and then just plug in the numbers. The acceleration is due to gravity and is equal to:

$a = g \sin \theta$

This gives the distance of:

$d = v_0 t + \frac{1}{2}at^2 = v_0 t + \frac{1}{2} g t^2 \sin \theta$

Solving this quadratic equation for $t$:

$t = \frac{-v_0 + \sqrt{ v_0^2 + 2g d \sin \theta }}{ g \sin \theta }$

We can only take the solution with a "+" sign since negative time makes no sense.

a)

$t_1 = \frac{\sqrt{2 \cdot 9.81 \text{ m/s}^2 \cdot 70 \text{ m} \cdot \sin(30^\circ) }}{ 9.81 \text{ m/s}^2 \cdot \sin(30^\circ) } = \boxed{ 5.34 \text{ s}}$

b)

$t_2 = \frac{-2.5 \text{ m/s} + \sqrt{ (-2.5 \text{ m/s})^2 + 2 \cdot 9.81 \text{ m/s}^2 \cdot 70 \text{ m} \cdot \sin(30^\circ) }}{ 9.81 \text{ m/s}^2 \cdot \sin(30^\circ) } = \boxed{ 4.86 \text{ s}}$

c) The answer isn't suprising, it's explained at the beggining of the task why the difference isn't big. In competitive events the difference in time comes to one hundreds of a second so every contribution matters.

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