I want to simulate a catapult throwing a rock in my computer game, but by design, I want all my units to shoot from a certain height, reach a maximum height and also hit a target that can be meters or kilometers away.
I tried solving the parabola needed for the equation using maths only since I was not able to do it with Kinematic formulas(I am not good at maths). For that, I solved the following system (Interactive here).
Where the parabola is constrained to pass through $x_0$ (launch point) and $x_2$ target point, and halfway the derivative of the parabola must be $0$ so that there is a maximum (or minimum).
The problem is with that solution I cannot fix the height where the maximum is going to be. However, If I fit a parabola by scaling in $x$ and in $y$ I can achieve the desired constraints.
Can anyone briefly explain, why is it possible to find a set of kinematic equations that satisfy a parabola, that a linear system can't find a solution for?
The goal here is to develop a method for determining the path of motion of a projectile which is launched from a point $(x_0, y_0)$ and hits a target at $(x_2, y_2)$. Because I want to derive the result, rather than trust some formula that I don't understand, I'm going to go through this from first principles. This system can be modeled as follows:
I am going to assume, for concreteness, that distance is measured in meters, and that time is measured in seconds. A projectile launched at time $t_0$ with initial velocity $(a,b)$ (that is, it is moving $a$ meters per second horizontally, and $b$ meters per second vertically) travels along a parabolic path given by $$ q(t) = \left( at + x_0, -\frac{1}{2}gt^2 + bt + y_0 \right), $$ where $q(t)$ is the projectile's position at time $t$, and $g$ is a gravitational constant (9.8 meters per second2 on Earth, but we can make it whatever we want, I suppose).
This ultimately comes from a differential equation. We know that the height, $y(t)$, satisfies $y''(t) = -g$, since the only force acting on the projectile is gravity. Then $$ \text{vertical velocity} = y'(t) = \int y''(t) = -gt + C, $$ where $C$ is a constant of integration. Since we know that the initial velocity is $b$, we have $b = y'(0) = C$, so $y'(t) = -gt + b$. Similarly, $$ \text{height} = y(t) = \int y'(t) = -\frac{1}{2}gt^2 + bt + C, $$ where $C$ is another constant of integration which, in this case, turns out to be $y_0$. This explains the $y$-coordinate of the position function $q$. Similar arguments give the $x$-coordinate, where we assume that no force is acting in the horizontal direction—the projectile just moves with a constant velocity of $x'(t) = a$.
Without loss of generality, suppose that the projectile is launched at time $t=0$, and hits its target at time $t_t$. By construction, $q(0) = (x_0, y_0)$. We also need $$ q(t_t) = \left( at_t + x_0, -\frac{1}{2} g t_t^2 + bt_t + y_0 \right) = (x_2, y_2). \tag{1}$$ Finally, we want the projectile to reach its maximum height at $x_1 = \frac{1}{2}(x_0 + x_2)$. While more elementary tools will get the job done, this happens when the vertical velocity is zero, i.e. when \begin{align} q'(t_m) &= \left( a, -gt_m + b \right) = \left( a, 0 \right), \tag{2} \end{align} where $t_m$ denotes the time at which the projectile reaches its maximum height. Putting this together, we have several quantities running around:
In a realistic setting, we aren't going to have much control over anything other than the initial velocity (for example, you can control how much powder you put in your cannon, but not much else). Then (1) and (2) imply that \begin{align} at_t + x_0 &= x_2, \tag{4}\\ -\tfrac{1}{2}gt_t^2 + bt_t + y_0 &= y_2, \tag{5}\\ -gt_m + b &= 0. \tag{6} \end{align} There are four unknowns here ($a$, $b$, $t_m$, and $t_t$), and three equations, so there is hope, but if we get any solution, there will be infinitely many more solutions—we will have some free parameter that we will have to choose.
Equations (4) and (6) give $$ \boxed{a = \frac{x_2-x_0}{t_t}} \qquad\text{and}\qquad \boxed{b = gt_m}. $$ To get a relation between $t_m$ and $t_t$, substitute $b=gt_m$ into (5) and solve for one of either $t_m$ or $t_t$ (you can go either way: solving for $t_m$ let's you determine when the projectile reaches its maximum height, while solving for $t_t$ let's you determine the time at which you hit your target). Getting $t_m$ in terms of $t_t$ is relatively straight-forward (and seems the natural thing to want to do): $$ -\frac{1}{2}gt_t^2 + gt_mt_t + y_0 = y_2 \implies \boxed{ t_m = \frac{y_2 - y_0 + \frac{1}{2}gt_t^2}{gt_t} }. \tag{7} $$ If you need to determine the maximum height from this, evaluate $q(t_m)$, and look at the $y$-coordinate. Here is a Desmos demonstration which models this situation—you can drag either the cannon or the target around, and use the slider to adjust the time-to-target.