Find the max projection angle so that distance from point of projection always increases We want to launch a projectile from the surface of earth so that its distance from the point of projection is always increasing. What is the maximum angle of projection for which this is possible?
One way to visualize this would be that there should not be a component of velocity opposite to the position vector of the particle, the point of projection being the origin, at all times of motion.
So following this line of thought I wrote :
$$\overrightarrow v \cdot \overrightarrow r > 0$$
Which did not yield a very nice expression in the angle $\theta$.
There should be better methods, as is often the case with physics, please suggest some of them.
 A: The equations of motions for a projectile are,
$$
x(t) = v_0 \cos(\theta)t,
$$
$$
y(t) = -\frac{1}{2}gt^2+v_0 \sin(\theta)t.
$$
Therefore the distance from the point of projection is,
$$
r(t)=\sqrt{x^2(t)+y^2(t)}.
$$
Since you want the distance from the point of projection is always increasing, we must have,
$$
\frac{dr(t)}{dt} > 0.
$$
By substituting $r(t)$, $x(t)$ and $y(t)$ in above equation, after some straightforward calculation you can easily obtain,
$$
\frac{dr(t)}{dt} = \frac{g^2 t^2 - 3 g t v_0 \sin(\theta) + 2 v_0^2}{\sqrt{g^2 t^2-4 g t v_0 \sin(\theta) + 4 v_0^2}}.
$$
Moreover, we know that if $a>0$, and $\Delta=b^2-4ac < 0$, then $at^2+bt+c > 0$ for all $t$. Therefore, we must have,
$$
a = g^2 > 0,
$$
$$
\Delta = g^2 v_0^2 \left(9 \sin^2(\theta) - 8\right) < 0,
$$
which results in
$$
\sin(\theta)<\sqrt{\frac{8}{9}} \to \theta < \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \approx 70.5288°
$$
A: Following up on my comment here is the (slightly simpler) math to compute this directly from the fact that product of gradients of two lines at right angles is -1:


A: 
This is the time where pv is perpendicular to velocity and if this time exists then distance of projectile from point of projection will not increase.
So take discriminant negative to vanish possibility of this time
