Consider a skier who is sliding without friction on the hill ${y = h(x)}$ in a two dimensional world. The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill. The second force automatically adjusts its magnitude so as to prevent the skier from burrowing into the hill. Suppose that the skier became airborne at some ${(x_0, y_0)}$ with ${y_0 = h(x_0)}$. How fast was the skier going?
My Method : Let mass of skier be ${m}$. Since only gravity is acting in the downward direction ,when the skier leaves the ground,from Newton's law which states that ${\vec{F}=m.\vec{a}}$, so $${\text{rate of change of velocity of skier}= \text{accelaration due to gravity}}.$$ Taking first derivative at ${(x_0,y_0)}$, we get $${\frac{dy_0}{dt}=h^{'}(x_0)\cdot\frac{dx_0}{dt}}.$$ Taking the second derivative of ${y=h(x)}$, we get $${a_y=h^{'}(x)\cdot{a_x} + h^{''}(x)\cdot\left({\frac{dx}{dt}}\right)^2}.$$ Now ${a_x = 0}$ since no force acts in the direction of x-axis and ${a_y=g}$ at the moment of going airborne. From the above equation we get $${v_{x_0}=\frac{dx}{dt}=\sqrt{\frac{g}{h^{''}(x_0)}}}$$ at ${x= x_0}$. Using this we get $${v_{y_0}=\frac{dy}{dt}=h^{'}(x) \cdot {\sqrt{\frac{g}{h^{''}(x_0)}}}}.$$ Thus the velocity of skier when he goes airborne is $${\large{\vec{v}=\sqrt{\frac{g}{h^{''}(x_0)}}\hat{i}}+ h^{'}(x).{\sqrt{\frac{g}{h^{''}(x_0)}}}\hat{j} }.$$ Now I am not sure whether my solution is correct or not. Even if it is correct I still don't feel like its a very good solution. Please help me improve my methodology and also in finding the flaws in the solution. Thank you! :)
Looks perfect to me. My only point of comment is that you should have $−h''(x_0)$ everywhere that you have $h''(x_0)$. The skier can only get airborn if he skies over a bump, meaning $h''(x_0)<0$.
I'm making the assumption that you've chosen the positive y-axis up.
Then, going airborn does not necessarily mean that he's accelerating in the direction of the positive y-axis. It suffices that the slope suddenly drops down.
If we take another look at your formula: $$a_y=h'(x)\cdot a_x+h''(x) \cdot \left(\!\frac {dx}{dt}\!\right)^2$$ We should have $a_y=-g$, which gives with $a_x=0$: $$\left(\!\frac {dx}{dt}\!\right)^2 = \frac{-g}{h''(x)}$$ $$\frac {dx}{dt} = \sqrt{\frac{g}{-h′′(x)}}$$
Note that if $h''(x) > 0$ this expression is undefined. Indeed in that case there is no chance for the skier to become airborn. It means that the slope goes down less and less, possibly ending in the horizontal.