Determine the velocity of the skater when they reach the point at the bottom of the halfpipe

Question

A skateboarder with a mass of $75$ kg rolls down a parabolic halfpipe (with shape described by the function $y = 0.2 x^2$). When the skateboarder is at a point $h_1 = 3$m above the ground, their velocity $v_1 = 3.0$ m/s. Neglecting friction, determine the velocity $v_2$ of the skater when they reach the point at the bottom of the halfpipe

My working out: $$mgh = mv^2/2$$ \begin{align} v & = \sqrt{2gh}\\ & = \sqrt{9.8 \times 3 \times 2} \\ & = 7.67 \text{m/s}\\ \end{align}

Is my working out and answer correct?

Answer 1

Your answer would be correct if the skateboarder started out stationary when they are 3m from the ground. But they aren't stationary - they have an initial speed of 3 m/s.

So your energy equation should be:

$\frac{1}{2}mv_1^2 + mgh = \frac{1}{2}mv_2^2$

You know $v_1=3$ and $h=3$ so you can solve to find the value of $v_2$.

Answer 2

When we know that:

$$\text{E}_{\space\text{initial}}=\text{E}_{\space\text{k}}+\text{E}_{\space\text{p}}=\frac{\text{m}_{\space\text{initial}}\cdot\text{v}_{\space\text{initial}}^2}{2}+\text{m}_{\space\text{initial}}\cdot\text{g}_{\space\text{initial}}\cdot\text{h}_{\space\text{initial}}\tag1$$

And when he is at the bottom:

$$\text{E}_{\space\text{final}}=\text{E}_{\space\text{k}}=\frac{\text{m}_{\space\text{final}}\cdot\text{v}_{\space\text{final}}^2}{2}\tag2$$

Now, when we know that:

• $$\text{g}_{\space\text{initial}}=\text{g}\tag3$$
• $$\text{m}_{\space\text{initial}}=\text{m}_{\space\text{final}}=3x\tag4$$
• $$\text{v}_{\space\text{initial}}=\text{h}_{\space\text{initial}}=x\tag5$$

So, you can write:

$$\frac{3x\cdot x^2}{2}+3x\cdot\text{g}\cdot x=\frac{3x\cdot\text{v}_{\space\text{final}}^2}{2}\space\Longleftrightarrow\space\text{v}_{\space\text{final}}=\pm\sqrt{x}\cdot\sqrt{x+2\cdot\text{g}}\tag6$$