# I Angle required to hit the target in projectile motion

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1. Apr 13, 2016

### Soren4

Consider the problem of projectile motion where the angle to hit a target $(x,y)$ is asked, once given the initial velocity magnitude $v_0$. The projectile is fired from the point $(0,0)$.

Here is the final formula for theta, solving the equations of motion (https://en.wikipedia.org/wiki/Traje...#Angle_required_to_hit_coordinate_.28x.2Cy.29)

$$\theta = \arctan{\left(\frac{v_0^2\pm\sqrt{v_0^4-g(gx^2+2yv_0^2)}}{gx}\right)} \tag{1}$$

Now if I impose $\Delta=v_0^4-g(gx^2+2yv_0^2)$ (for which we have the only solution $\theta=\arctan \frac{v_0^2}{gx}$) I find $$y=\frac{v_0^2}{2g}-\frac{g}{2v_0^2}x^2\tag{2}$$

Now in my view this is the equation of a parabola which describes the points $(x,y)$ in plane that can be hit only firing at $\theta=\arctan \frac{v_0^2}{gx}$, once given the initial velocity $v_0$.

On my textbook it is claimed that $(2)$ represents the equation of the trajectory of the projectile, if fired at $\theta=\arctan \frac{v_0^2}{gx}$.

I don't think that this is possible, to begin with the fact that $(2)$ does not pass through $(0,0)$.

Can anyone tell me what exactly $(2)$ means?

2. Apr 13, 2016

### PeroK

(2) is a relationship between x, y, g and v that results in a single solution for theta.

3. Apr 13, 2016

### Soren4

Thanks for the reply! Ok that's a relation between $x,y,v$. Is it correct to see it as the locus of points in plane that, given the particular value of $v$, can be reached iff the projectile is fired at the angle $\theta=\arctan \frac{v_0^2}{gx}$? For those points there is infact this one only solution in $(1)$.