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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.

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  • $\begingroup$ So you have the parabola, right? Consider the starting point of the parabola. Now, the question asks: what is the initial angle of this parabola? How would you calculate this? Any ideas? The angle we want is the angle between the ground ($x$-axis) and the graph of the parabola. $\endgroup$ – Matti P. Feb 11 at 10:42
  • $\begingroup$ What is the equation of your parabolic function? $\endgroup$ – Max Feb 11 at 10:48
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    $\begingroup$ Are we on the Earth or on the Moon ? To calculate the initial speed you have to implicitly make an assumption about the acceleration due to gravity, or explicitly include $g$ in your soilution. $\endgroup$ – gandalf61 Feb 11 at 12:10
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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 $\frac{-g}{2v_0^2cos^2\alpha}=-\frac{49}{2560}$ $\Rightarrow, v_0 =\sqrt{\frac{2560g}{98\cos^2\alpha}}=20m.s^{-1}$

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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.

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  • $\begingroup$ I think you should have $\cos^2 \theta$ instead of $\cos \theta$ in the trajectory equation. $\endgroup$ – gandalf61 Feb 11 at 12:08
  • $\begingroup$ @gandalf61 I corrected it. $\endgroup$ – Paras Khosla Feb 11 at 12:42

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