Find constant acceleration needed to reach point (possibly related to projectile motion on slopes) [closed]

Question

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.

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.

Answer 1

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

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UPDATE in response to your attempts at a solution :

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

$u$, $\beta$ 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 $T$.