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Given a particle with a position $p_0$ and an initial velocity $\vec{v_0}$, what acceleration $\vec{a}$ do we need to reach point $p_1$ and how long until we have reached $p_1$? The magnitude of the acceleration must be $\leq A_{max}$, and the time to reach the $p_1$ should be minimal. The acceleration should be constant, i.e. the same acceleration vector should be used for the entire trip.

Our first attempt was to use the basic kinematics formula from Wikipedia, but because both $t$ and $A$ are unknown, this proved unfruitful.$$P_1=\frac12at^2+v_0t+P_0$$

Edit: Reformulation of problem in the form of projectile motion on slopes and attempt at solution using the link provided by sammy gerbil.

Drawing of the problem

Problem: A projectile is launched up/down a slope from $O$ with velocity $\vec{u}$. The projectile lands at $P$. The angle between the slope and $\vec{u}$ is $\beta$. The force of gravity is $g$. What is the angle $\alpha$ (i.e. the angle between the slope and the horizontal) that minimizes the time required to reach $P$? In other words, in which direction should gravity be applied to minimize flight time?

Attempted solutions: We tried using the formula for range of flight from the link sammy provided $$T = \frac{2\|\vec{u}\|\sin\beta}{g\cos\alpha}$$ Our thought is that minimizing that function should do the trick. We don't really know how though.

We also tried to calculate $\alpha$ using the many different formulae on the same link, but there seems to be too many variables. For example, the formula for range of flight has both $\alpha$ and $2\theta - \alpha$ as unknowns.

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  • $\begingroup$ well, if v_0 is non-zero, and it's direction is towards p_1, the acceleration can take many values and you'd still reach p_1... The time it would take depends on the the acceleration you pick $\endgroup$ – Mr.WorshipMe May 9 '16 at 12:39
  • $\begingroup$ We want to minimize the time $t$. $\endgroup$ – Hampus Fristedt May 9 '16 at 12:39
  • $\begingroup$ Do you need to stop at p_1? $\endgroup$ – Mr.WorshipMe May 9 '16 at 12:40
  • $\begingroup$ No, we do not need to stop at $p_1$. $\endgroup$ – Hampus Fristedt May 9 '16 at 12:41
  • $\begingroup$ So choose $a=A_{max}$.. $\endgroup$ – Mr.WorshipMe May 9 '16 at 12:42
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This is the same as the problem of a projectile launched on an inclined plane. The link below shows you how to find range and time of flight.

You will need to adapt your problem to fit that described. O is the point of launch (A) and P is the target (B). OP=AB is the range $R$ which you know. In this scheme acceleration is $ g=A $ along the -y axis.

I think the choice $g=A_{max}$ will always give you the shortest time; you just need to find the correct angle $\alpha$ in which to apply the acceleration $g$ in order to land on the target P. I shall leave you to work out the details.

https://cnx.org/contents/--TzKjCB@8/Projectile-motion-on-an-inclin


UPDATE in response to your attempts at a solution :

We are using co-ordinate system 2b from the link : axes are horizontal and vertical.

enter image description here

$u$, $\beta$ and the range OP = $R$ are given in your problem. The acceleration (gravity $g=A$) acts vertically down. We can vary the magnitude $g$ and direction angle $\alpha$ of the acceleration in order to minimise the time of flight $T$.

I suggest that $g=A_{max}$ always gives the smallest value of $T$, but I do not see how to prove it rigorously. It is clearly true when $\beta$ = 0 or 180 degrees. So the problem boils down to finding the corresponding value of $\alpha$.

As you state, the time of flight is given by
$T =\frac{2usin(\beta)}{gcos(\alpha)}$.
Using $g=A_{max}$, angle $\alpha$ is constrained by the requirement that the projectile must reach the target P which is distance $R$ from O where
$R = \frac{u^2}{g cos^2(\alpha)} (sin(2\theta-\alpha) - sin(\alpha)) $
and $\theta=\alpha+\beta$.

So it is first necessary to input $R, u, g, \theta$ into this eqn and solve for $\alpha$, then substitute into the eqn for $T$. It is not necessary to minimise any function, but solving to find $\alpha$ is difficult.

Using the Sum-to-Product formula we have
$sin(2\theta-\alpha) - sin(\alpha) = 2 cos(\alpha+\beta) sin(\beta) $.
Then
$2cos^2(\alpha) = 1+cos(2\alpha) = k cos(\alpha+\beta)$
where $k = \frac{4u^2sin(\beta)}{gR}$.

Use the known values to get a value for $k$, then plot $y=1+cos(2\alpha)$ and $y=k cos(\alpha+\beta)$ against $x=\alpha$. Points of intersection give approximate solutions for $\alpha$ which you can make arbitrarily accurate by some numerical method (eg Newton-Raphson, or simple 'trial and improvement'). Then substitute to find $T$.

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  • $\begingroup$ Thank you very much! This helps but we still can't figure it out. We edited our original question. Could you please help? $\endgroup$ – Hampus Fristedt May 13 '16 at 10:54
  • $\begingroup$ @HampusFristedt : One thing I am uncertain about : the initial velocity $\vec{v_0}$ (which I have called u) seems to have a fixed magnitude, but does it also have a fixed direction? Or is the direction variable ("arbitrary" as you describe it)? $\endgroup$ – sammy gerbil May 13 '16 at 13:10
  • $\begingroup$ The initial velocity is fixed both in magnitude and direction. $\endgroup$ – Hampus Fristedt May 13 '16 at 13:11
  • $\begingroup$ @HampusFristedt : This is more difficult than I expected! I am still working on it but I will try to give you an answer within 24 hours. $\endgroup$ – sammy gerbil May 13 '16 at 15:16
  • $\begingroup$ Yes, I have also found it to be a deceptively difficult problem. No need to rush; I very much appreciate you helping regardless! $\endgroup$ – Hampus Fristedt May 13 '16 at 15:44

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