Projectile motion, initial angle and velocity Question: A projectile motion passes $(16i+7.1j)$, $(0i+0j)$ and $(32i+4.4j)$. Determine the initial speed and velocity. (At the origin) 
My attempt: I converted the $i$ and $j$ in $x$ and $y$ to find the parabolic cartesian formula for the projectile. However, while this helped determine the angle of projection, it didn’t show me the initial velocity, so I’m stuck. 
 A: If $v_0$ is the initial velocity, $g$ the acceleration, and $\alpha$ the initial angle ($\alpha=( \vec{i},  \vec{v_0}))$ then $ y=\frac{-g}{2v_0^2cos^2\alpha}X^2+X\tan\alpha$. You can deduce $\tan\alpha=\frac{3}{4}$ and  $\frac{-g}{2v_0^2cos^2\alpha}=-\frac{49}{2560}$ $\Rightarrow, v_0 =\sqrt{\frac{2560g}{98\cos^2\alpha}}=20m.s^{-1}$
A: Hint: Note that the trajectory of a projectile in the $xy$ plane is given by the following equation which can be proven simply by performing some manipulations on the equations of position in the $x$ and $y$ directions. $$y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta} \tag1$$Observe that you have two ordered pairs of positional coordinates, $\begin{bmatrix} 16 \\ 7.1\end{bmatrix}$ and $\begin{bmatrix}32\\4.4 \end{bmatrix}$. Plugging in those values would give you two equations with $u$, the magnitude of initial velocity and $\theta$, the angle of projection as unknowns solving which you can determine the inital velocity.
Aliter: Another approach would be to use the relation between the angle of projection, $\theta$ and the initial speed, $u$. This is given by $$\theta=\arctan\biggl(\dfrac{u^2 \pm\sqrt{u^4-g(gx^2+2yu^2)}}{gx}\biggr) \tag2$$Note however that this is the same formula as above wherein $\theta$ is rewritten as an explicit function of $u$. To prove this simply write $\cos\theta$ in terms of $\tan\theta$ in equation $(1)$, make up a quadratic equation with $\tan\theta$ as the unknown and solve. At the very end take arctangent on both sides of the equation and you have the relation.
