# 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.

• 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. – Matti P. Feb 11 at 10:42
• What is the equation of your parabolic function? – Max Feb 11 at 10:48
• 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. – gandalf61 Feb 11 at 12:10

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 $$\frac{-g}{2v_0^2cos^2\alpha}=-\frac{49}{2560}$$ $$\Rightarrow, v_0 =\sqrt{\frac{2560g}{98\cos^2\alpha}}=20m.s^{-1}$$

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.

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