determining the start velocity and angle based on the goals(End point) angle and coordinates. ========
I'm trying to figure out the start angle and velocity of an parabola, When i know only the start coordinates and the end point coordinates and the angle which it reaches the endpoint.
I'm tasked with figuring this out by using derivative method.
So the known parameters are:
Start point (0,0)
end point (6,4)
end point angle (-18)
unknown are:
velocity = unknown.
start angle =unkown.
This is a homework assiment thats been bothering me for months now.
Here is a picture of what i've been working with:



The equation of motion of a projectile starting from the origin, and shot at an angle $\theta $ above the horizon, at an initial velocity $\mathbf{v_0}$ is
$\mathbf{P}(t) = \mathbf{v_0} t + \dfrac{1}{2} \mathbf{g} t^2 $
Now, $\mathbf{P}(t) = ( x(t), y(t) ) $ and $\mathbf{v_0} = v_0 (\cos \theta, \sin \theta) $, and $ \mathbf{g} = (0, - g ) $ , where $g = 9.81 \ \text{m/s}^2$
Hence,
$ x(t) = v_0 \ cos (\theta) \ t  $
$ y(t) = v_0 \ \sin(\theta) \ t - \dfrac{1}{2} \ g \ t^2 $
Solving for $t$ from the first equation for $x(t)$,
$ t = \dfrac{x}{v_0 \cos \theta} $
Substitute this into the equation for $y(t)$
$ y = \tan(\theta) \ x - \dfrac{g \ x^2}{2 v_0^2 \cos^2(\theta)  }\hspace{25pt} (1) $
Substitute the point $(6, 4)$ into equation $(1)$
$ 4 = 6 \tan(\theta) - \dfrac{ 18 g }{ v_0^2 \cos^2 \theta } \hspace{25pt} (2)$
Differentiate equation $(1)$ with respect to $x$ to get the slope of the tangent to the projectile trajectory,
$ \dfrac{dy}{dx} = \tan(\phi) = \tan(\theta) - \dfrac{ g x }{ v_0^2 \cos^2 \theta } \hspace{25pt} (3) $
Since at the end point (which is $(6, 4)$) the angle is $\phi = -18^\circ$, then substituting this into $(3)$, gives us
$ \tan(-18^\circ) = \tan(\theta) - \dfrac{ 6 g }{ v_0^2 \cos^2 \theta } \hspace{25pt}(4)$
Eliminating the second term from equations $(2)$ and $(4)$, we get
$ 4 - 3 \tan(-18^\circ) = 3 \tan(\theta) $
Hence, $ \theta = \tan^{-1}\bigg( \dfrac{ 4 - 3 \tan(-18^\circ) }{3} \bigg) = 58.9^\circ$
Substitute this into $(2)$ and solve for $v_0$,
$ v_0 = \sqrt{ \dfrac{18 g }{ (6 \tan(\theta) - 4) \cos^2(\theta) } } = 10.55 \ \text{m/s}$
