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/Trajectory_of_a_projectile#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 $\Delta=0$ [Edit: $\Delta=v_0^4-g(gx^2+2yv_0^2)$ is the discriminant (https://en.wikipedia.org/wiki/Discriminant)] (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?

  • $\begingroup$ What's $\Delta$? $\endgroup$
    – David Z
    Apr 13, 2016 at 13:20
  • $\begingroup$ How did you arrive at $(2)$? It seeems to me you cancelled the target condition $x$ (which is a fixed number for any given throw) in $\theta = \arctan(\frac{v^2}{gx})$ against the variable $x$ in the expression for $y(x)$, that doesn't work. Better call the target condition $x_1$ or something. $\endgroup$
    – ACuriousMind
    Apr 13, 2016 at 13:41
  • $\begingroup$ @ACuriousMind I got $(2)$ simply imposing $\Delta =0$ (to see in which conditions one only solution exists for the problem), I did not susbstitute anywhere. The thing you said is exaclty what my book does: substitute $\theta=\arctan \frac{v^2}{gx}$ in the trajectory equation $y(x)$ and get $(2)$, canceling the $x$ (which is wrong to me too). $\endgroup$
    – Sørën
    Apr 13, 2016 at 13:52

2 Answers 2


If your textbook actually derives $(2)$ as the motion of a thrown object, throw it away.

The general trajectory of an object thrown from $(0,0)$ at angle $\theta$ is $$ y(x) = x\tan(\theta) - \frac{gx^2}{2v^2}(1+\tan(\theta)^2)\tag{1}$$ and now you say you "impose $\Delta = 0$". Let's analyse that "imposing" a bit more carefully, it is equivalent to $$ v^4 = g(gx^2 + 2yv^2)$$ which, in turn, gives $$ y(x) = \frac{v^2}{2g} - g\frac{x^2}{2v^2}\tag{2}$$ just by solving for $y$ (not by inserting into $(1)$). If it were valid to "impose $\Delta = 0$", then this would have to be compatible with the trajectory $(1)$, but as you already noticed, plugging in $x=0$ gives $y= 0 $ for $(1)$ and $y=\frac{v^2}{2g}$ for $(2)$, so the only possible trajectory with $\Delta = 0$ is the one with $v=0$ where the object doesn't move at all. It's simply invalid to expect the particular case $\Delta = 0$ to give any non-trivial solution.

  • $\begingroup$ Thanks for the answer, so $(2)$ is not a trajectory for sure. If I may ask is it correct to see $(2)$ as I stated in the question? i.e. the equation of a parabola which describes the points (x,y) in plane that can be hit only firing at one angle, $θ=\arctan\frac{v_0^2}{gx}$, once given the initial velocity $v_0$. Imposing $\Delta=0$ is to see for what points there is only one angle for reach the target (one solution only in $(1)$), so I think that's the correct interpretation for $(2)$. $\endgroup$
    – Sørën
    Apr 13, 2016 at 19:29
  • 1
    $\begingroup$ @Sørën: No, $\Delta = 0$ is simply inconsistent with the assumption of firing something from $(0,0)$ with non-zero velocity. It doesn't have any meaning. There is never just one angle to reach any given point - two points do not define a parabola. $\endgroup$
    – ACuriousMind
    Apr 13, 2016 at 19:38
  • $\begingroup$ @ACuriousMind Old thread, I know. Just wondering, can we actually put $v$ = $0$ in equation ($2$) ? Won't the second term on the RHS become $\frac{0}{0}$ ? I understand we put $x$ = $0$ in the numerator, so that term should become zero even before we put a value of $v$, because the projectile is being launched from ($0$,$0$). But still, can we put $v$ = $0$ in equation ($2$)? I'm confused $\endgroup$
    – 4d_
    Feb 15, 2019 at 7:54

Equation (2) causing the confusion is just the bounding parabola: the limitation of points which could be hit (setting the angle of trajectory to 45° which gives the furthest limits of possible targets and also is the set of points which have only one angle to reach them (contrary to the false statement by Soren).


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