# Problem of minimum time for a falling mass.

A mass falls from a point A (height h) to a point P (height $$0$$) and then it continues to move with the speed acquired in P. I have to find the position of the point P on the ground in order to have the minimum time. I take x the lenght on the ground of A from P.

$$E_A=mgh$$ (the mass falls down and it has initial speed $$0$$).

$$E_P= \frac{1}{2}m (v_P)^2$$ (P is on the ground amd I've taken the ground as my reference plane).

$$E_A=E_P \Rightarrow (v_P)^2=2gh \Rightarrow v_P= \sqrt{2gh}$$.

This velocity remains constant until the mass reaches the final point B.

$$v_P=v_A+gt_1 \Rightarrow t_1= \frac{v_P-v_A}{g}= \frac{\sqrt{2gh}-0}{g}=\sqrt{\frac{2h}{g}}$$

$$t_2= \frac{L-x}{\sqrt{2gh}}$$

$$t_{tot}=t_1+t_2= \sqrt{\frac{2h}{g}} + \frac{L-x}{\sqrt{2gh}} = f(x)$$

But $$\frac{df}{dx}=0$$ is never verified and I don't know how to procede.

In the solution on the book I don't understand why it says that $$x$$ is the minimum of $$L$$ and $$\frac{h}{\sqrt 3}$$

and it analyze the cases of $$h \le \sqrt3 l$$ and of $$h > \sqrt3 l$$.

The issue is in your equation $$v_P=v_A+gt_1$$ The acceleration is not $$g$$, but the component of $$g$$ along the $$AP$$ line. You need to decompose $$g$$ into components parallel and perpendicular to $$AP$$. If $$\alpha=\angle APC$$, then $$g_{||}=g\sin\alpha$$ From $$\triangle APC$$, $$\sin\alpha=\frac h{\sqrt{h^2+x^2}}$$ Can you continue from here?

• ok I've found the minimum for $x= \frac{h}{\sqrt3}$
– Anne
Nov 14 '20 at 16:31
• Note that you need to consider the case when $\frac h{\sqrt 3}>L$ Nov 14 '20 at 16:32
• but I don't understand how to calculate $t_{min}$ in the case of $L /ge \frac{h}{\sqrt 3}$ and $L < \frac{h}{\sqrt 3}$
– Anne
Nov 14 '20 at 16:39
• When $L\ge h/sqrt 3$ you have $x=h/sqrt 3$, so just plug it into your formula for $t_{tot}$. When $L<h/sqrt 3$, points $P$ and $B$ are the same. You can then calculate the acceleration as $gh/\sqrt{h^2+L^2}$, and the distance traveled is $\sqrt{h^2+L^2}$. Nov 14 '20 at 16:46
• yes, indeed, now I see, it's all clear thank you very much!
– Anne
Nov 14 '20 at 16:46