# Finding the initial launch angle of a projectile so that it reaches a certain point when a 2D acceleration is applied.

I'm trying to write an AI that can predict at which angle to throw a projectile so that it reaches a certain target position. It is on a 2D plane where the only acceleration applied to the object is wind (no gravity).

What I know beforehand is:

• The launch velocity. $$v_i$$
• The starting and end points.
• The wind's acceleration vector.
• $$W_x, W_y$$

Looking back at high school physics I wind up with these equations.

$$X = v_i * cos(\theta) * \Delta t + \frac{1}{2} * W_x * \Delta t^2$$ $$Y = v_i * sin(\theta) * \Delta t + \frac{1}{2} * W_y * \Delta t^2$$

(the equations above assume a starting position of (0,0))

The issue is that I can't seem to figure out how to solve this equation since there are two unknowns (theta and the time to reach the final position). When only factoring in gravity this is easily solvable but since there is an acceleration on the X-axis it's more complicated.

The closest I've gotten is using the quadratic formula to get an equation that solves for time, and then just brute-forcing possible angles until the times match up for both equations. I'm wondering if there's a more elegant algorithm to use instead.

## 3 Answers

Given the following real input parameters:

$$\left(x_i,\,y_i\right), \quad \quad \quad \left(x_f,\,y_f\right), \quad \quad \quad \left(a_x,\,a_y\right), \quad \quad \quad v_i > 0$$

and define the differences:

$$\Delta x := x_f - x_i\,, \quad \quad \quad \Delta y := y_f - y_i$$

we need to solve the system of equations:

$$\begin{cases} \Delta x = v_i\cos\theta\,\Delta t + a_x\,\Delta t^2/2 \\ \Delta y = v_i\sin\theta\,\Delta t + a_y\,\Delta t^2/2 \\ \end{cases}$$

where solutions $$\Delta t > 0$$ and $$0 \le \theta < 2\pi$$ are of interest.

So, rewriting the system of equations:

$$\begin{cases} \frac{2\,\Delta x - a_x\,\Delta t^2}{2\,v_i\,\Delta t} = \cos\theta \\ \frac{2\,\Delta y - a_y\,\Delta t^2}{2\,v_i\,\Delta t} = \sin\theta \\ \end{cases}$$

it follows that:

$$\left(\frac{2\,\Delta x - a_x\,\Delta t^2}{2\,v_i\,\Delta t}\right)^2 + \left(\frac{2\,\Delta y - a_y\,\Delta t^2}{2\,v_i\,\Delta t}\right)^2 = 1$$

that is, by reworking:

$$\left(a_x^2 + a_y^2\right)\Delta t^4 - 4\left(v_i^2 + a_x\,\Delta x + a_y\,\Delta y\right)\Delta t^2 + 4\left(\Delta x^2 + \Delta y^2\right) = 0$$

where, defined the parameters:

$$a := a_x^2 + a_y^2\,, \quad \quad \quad b := - 4\left(v_i^2 + a_x\,\Delta x + a_y\,\Delta y\right), \quad \quad \quad c := 4\left(\Delta x^2 + \Delta y^2\right)$$

it's clear that there are solutions $$\Delta t > 0$$ if and only if:

$$a + c \ne 0 \quad \quad \quad \text{and} \quad \quad \quad b < 0 \quad \quad \quad \text{and} \quad \quad \quad b^2 - 4\,a\,c \ge 0$$

and in this case it's only one if:

• $$a = 0$$ from which $$\Delta t = \sqrt{\frac{c}{-b}}$$;

• $$c = 0$$ from which $$\Delta t = \sqrt{\frac{-b}{a}}$$;

• $$b^2 - 4\,a\,c = 0$$ from which $$\Delta t = \sqrt{\frac{-b}{2\,a}}$$;

otherwise are two:

$$\Delta t_1 = \sqrt{\frac{- b - \sqrt{b^2 - 4\,a\,c}}{2\,a}} < \Delta t_2 = \sqrt{\frac{- b + \sqrt{b^2 - 4\,a\,c}}{2\,a}}\,.$$

In particular, if there are solutions $$\Delta t > 0$$, the initial velocity vector is trivially determined:

$$v_{x,i} = v_i\,\cos\theta = \frac{\Delta x}{\Delta t} - \frac{a_x}{2}\,\Delta t\,, \quad \quad \quad v_{y,i} = v_i\,\sin\theta = \frac{\Delta y}{\Delta t} - \frac{a_y}{2}\,\Delta t$$

and consequently also the angle measured with respect to the positive semi-axis of the abscissa:

$$\theta = \begin{cases} \pi + \arctan\left(v_{y,i}/v_{x,i}\right) & \text{if} \; v_{x,i} < 0 \\ 3\pi/2 & \text{if} \; v_{x,i} = 0 \, \land \, v_{y,i} < 0 \\ \pi/2 & \text{if} \; v_{x,i} = 0 \, \land \, v_{y,i} > 0 \\ 2\pi + \arctan\left(v_{y,i}/v_{x,i}\right) & \text{if} \; v_{x,i} > 0 \, \land v_{y,i} < 0 \\ \arctan\left(v_{y,i}/v_{x,i}\right) & \text{if} \; v_{x,i} > 0 \, \land v_{y,i} \ge 0 \\ \end{cases}$$

which, if desired, is equivalent to writing:

$$\theta = \begin{cases} 2\pi + \text{arctan2}\left(v_{y,i},\,v_{x,\,i}\right) & \text{if} \; v_{y,i} < 0 \\ \text{arctan2}\left(v_{y,i},\,v_{x,\,i}\right) & \text{if} \; v_{y,i} \ge 0 \\ \end{cases}$$

where $$\text{arctan2}(y,\,x)$$ associates an angle in the interval $$(-\pi,\,\pi]$$ for each point $$(x,\,y) \ne (0,\,0)$$.

Since this is a uniformly accelerated motion, both in the horizontal and vertical directions, the parabolic trajectories will generally be rototranslated in the plane. For example, assuming:

$$\left(x_i,\,y_i\right) = (1,\,1), \quad \quad \quad \left(x_f,\,y_f\right) = (3,\,2), \quad \quad \quad \left(a_x,\,a_y\right) = (1,\,-1), \quad \quad \quad v_i = 2$$

the two trajectories solution of the problem are:

where blue is for $$\Delta t_1 \approx 1.06$$ and $$\theta_1 \approx 47.4°$$, while red is for $$\Delta t_2 \approx 2.98$$ and $$\theta_2 \approx 114°$$.

Finally, it isn't difficult to prove that the area enclosed by the two parabolic arcs is equal to:

$$\mathcal{A} = \frac{\left|a_x\,\Delta y - a_y\,\Delta x\right|\left(\Delta t_2^2 - \Delta t_1^2\right)}{12} = \frac{\sqrt{15}}{2} \approx 1.94\,.$$

I hope it's clear enough, good luck! ^_^

One approach is to solve the first equation for $$\cos \theta$$ and the second equation for $$\sin \theta$$, and then substitute both into $$\sin^2 \theta + \cos^2 \theta = 1$$. Another approach is to substitute both into $$\frac{\sin \theta}{\cos \theta} = \tan \theta$$. I'm not sure which approach yields simpler algebra. Either is preferable to dealing with a mess of trig and inverse trig functions.

Let's break the trajectory into two parts, ascent and descent. Ascent is till the body reached its maximum height. So let's see what the max height and time of time as ascent should be.

The projection of motion on vertical axis,

Initial velocity, $$u_i$$ is,

$$u_i = v_i sin \theta$$

Final velocity in ascent is $$0$$, therefore the time of ascent $$t_a$$ is,

$$t_a = \frac{v_i sin\theta} {W_y}$$

Assuming level terrain, the thrown object now will cover this height downward to fall on the ground. The time of descent is also the same as $$t_a$$ (if your wind acceleration $$W_y$$ is constant as we have treated during ascent, which prolly isn't true as the acceleration may itself depend on the speed).

So the distance covered in the horizontal direction (range) is what we are after,

$$X = v_i cos\theta (2t_a) + \frac{1}{2} W_x (4t_a^2)$$.