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enter image description hereA 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 $.

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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?

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  • $\begingroup$ ok I've found the minimum for $x= \frac{h}{\sqrt3}$ $\endgroup$
    – Anne
    Nov 14 '20 at 16:31
  • $\begingroup$ Note that you need to consider the case when $\frac h{\sqrt 3}>L$ $\endgroup$
    – Andrei
    Nov 14 '20 at 16:32
  • $\begingroup$ but I don't understand how to calculate $t_{min}$ in the case of $L /ge \frac{h}{\sqrt 3}$ and $L < \frac{h}{\sqrt 3}$ $\endgroup$
    – Anne
    Nov 14 '20 at 16:39
  • $\begingroup$ 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 $\sqrt{h^2+L^2}$. $\endgroup$
    – Andrei
    Nov 14 '20 at 16:46
  • $\begingroup$ yes, indeed, now I see, it's all clear thank you very much! $\endgroup$
    – Anne
    Nov 14 '20 at 16:46

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