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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.

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  • $\begingroup$ What do you mean exactly by "point of projection is always increasing"? Are you simply looking for the angle of projection which yields the maximum range? In that case you will most definitely find your answer here: en.wikipedia.org/wiki/Range_of_a_projectile $\endgroup$ – Phonon Aug 2 '14 at 10:34
  • $\begingroup$ Hi Shubham. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Aug 2 '14 at 10:41
  • $\begingroup$ No, my question is different. Suppose we launch a projectile at a very steep angle. During its motion, the distance from point of projection( that would simply be the magnitude of its displacement at that time), would increase till a certain time and then start decreasing. You can easily see this by drawing the path and drawing straight lines between point of projection different points on its path, a downward parabola. $\endgroup$ – Shubham Aug 2 '14 at 10:45
  • $\begingroup$ @Qmechanic, this is actually a homework question from my physics book. Yes, I have read the SE policy on homework like questions. Even I do not expect a full solution. $\endgroup$ – Shubham Aug 2 '14 at 10:48
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    $\begingroup$ Another way to look at this - the "limit" trajectory is the one where the path of the projectile is at right angles to the vector from the origin to the vector. $\endgroup$ – Floris Aug 2 '14 at 12:36
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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

$$ \sin(\theta)<\sqrt{\frac{8}{9}} \to \theta < \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \approx 70.5288° $$

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  • $\begingroup$ I was missing just the last part, this was the same not so nice expression I had arrived at(Initially I had tried this approach). Just did not know how to proceed. $\endgroup$ – Shubham Aug 2 '14 at 14:02
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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:

enter image description here enter image description here

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    $\begingroup$ That's exactly how my teacher solved it! This approach looks nicer from a physicist's perspective. $\endgroup$ – Shubham Aug 2 '14 at 18:01
  • $\begingroup$ it is very good assuming such a trajectory exists. But conveniently, the values for the angle and t are within the allowed range so it works. The method can't be extended if it just happened that such a trajectory does not exist, because we can't make a relationship between angle from the normal and velocity using the cross product because the cross product only tells you how close they are to right angles, not how large the difference between angles is $\endgroup$ – lucky-guess May 27 '17 at 12:36
  • $\begingroup$ sorry if that didn't make sense $\endgroup$ – lucky-guess May 27 '17 at 12:41
  • $\begingroup$ I am using the dot product not the cross product. Yes if there is no solution there is no solution... I don't understand your comment sorry. $\endgroup$ – Floris May 27 '17 at 13:03
  • $\begingroup$ If you never get to right angle, then the point where the projectile lands is the furthest point... that is not a solution to the above equation but it would be easy enough to find. $\endgroup$ – Floris Oct 15 '17 at 19:20

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