- #1

- 128

- 2

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?