Projectile on Inclined Plane: Finding Angle of Inclination

[jal3ous]

Homework Statement

A projectile is thrown from a point A on an inclined plane field of unknown slope α, and hits the field on a point B. We know the angle between the initial velocity and the field: θ. We know the magnitude of the velocity: V. We know the the acceleration of gravity: g. We know the slant range of the projectile (distance between A and B): R. We have no friction.

Calculate α in terms of θ,V,g, and R.

I worked on this problem and I always get a complex 4th degree polynomial equation or a trigonometric equation that I can't figure out how to solve.

Homework Equations

V sin(alpha + theta)t = R sin(alpha)
V cos (alpha+theta)t + (1/2)gt^2 = R cos(alpha)

The Attempt at a Solution

Consider alpha the angle between the gravity vector and the field. Consider a 2D reference frame (X,Y) that contains the trajectory and let Y be vertical and opposite to g and X be horizontal and such that the trajectory is in X > 0. if we decompose the equation of motion along X we get: V sin(alpha + theta)t = R sin(alpha). along Y we get: V cos (alpha+theta)t + (1/2)gt^2 = R cos(alpha). if we divide the second equation by the first we get a trigonometric equation with cot(alpha+theta), sin(alpha+theta), and cot(alpha). I couldn't go any further from here

 

[BvU]

Hi jal, welcome (a bit belated) to PF !

You have made an unconventional choice for $\alpha$ ! Usually we say $\alpha = 0$ means horizontal. For you it's straight down. Never mind.

$\theta=0$ means parallel with the plane and I can't find a flaw in your story. Dividing 2nd by 1st still leaves a t which you eliminate with the first eqn. Leaving not only the ones you mention, but another $\sin\alpha$ as well.

It is a single equation with a single unknown. I don't see a reasonable straightforward way towards an expression like $\alpha = ...$ (with no $\alpha$ on the RHS). I would solve it numerically.


And I do wonder if just this single equation (with $\alpha$ on both sides) isn't enough answer for the composer of the exercise.

Funny how such a simple situation can be made so complicated. After all, it's just a parabola passing to a suitable origin (point A) with an equation like y = P x ( x - Q) . Two degrees of freedom, three givens, so room for pinning down $\alpha$.

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Now we follow Tuesdays advice by @A.T. from post #2 in this thread (which is a duplicate of the current thread -- PF frowns on that ! ) and decouple $\theta$ and $\alpha$ at the (small) cost of having uniformly accelerated motion in two directions instead of one:

We say (going back to the more conventional $\alpha = 0$ and $\theta = 0$ when horizontal and I also like g = +9.81 m/s², sorry) $$
x = v_0\cos\theta\; t - {\tfrac 1 2} g \sin\alpha\; t^2 \\
y = v_0\sin\theta\; t - {\tfrac 1 2} g \cos\alpha\; t^2 \\
$$and know that $y=0$ at $t=0$ and at $$
t={2v_o\sin\theta\over g \cos\alpha}
$$You get somewhat cleaner equations, easier to type in when solving with e.g. excel :

But I don't get much further than an equation like $P\sin\alpha - Q\cos\alpha = C \cos^2\alpha$ (with P, Q and C known) which for me is too difficult

 

[jal3ous]

Thank you BvU for your reply!
Not a complete solution but very useful :)
I'll keep trying based on what u propose and see if I can get any further...