Deriving smallest possible launching angle such target hits a point from geometrical envelope

Question

I'm trying to understand fact-8 in page -9 of Jaan Kalda's PDF,

Fact 8: If target is same level as cannon, then the optimal launching angle (corresponding to smallest launching speed is 45 deg)

Below, it is said that we can find this from the equation of envelope of parabola equation (this is what was involved in pr.19), so I derived as follows.

I begin with time independent equation of projectile motion:

$$ y = x \tan \theta - \frac{g}{2} ( \frac{x \sec \theta}{u})^2 \tag{1}$$

Now, to find condition for enevelope, I need to have the condition that $ \frac{\partial y}{\partial \theta}=0$ i.e: find the condition on $\theta$ such that $y$ is maximized for a given $x,u$:

$$ 0 = x \sec^2 \theta - \frac{gx^2}{u^2} \sec^2 \theta \tan \theta$$

Or,

$$ \frac{u^2}{g \tan \theta} = x \tag{2}$$

Now, suppose that projectile equations are written about the place where we shoot projectile as $(0,0)$ then it must be that target is at vertical height of 0, pluggin this in $(1)$

$$ 0 = x \tan \theta - \frac{g}{2} ( \frac{x}{u})^2 \sec^2 \theta$$

PLuggying $2$ and simplfying, I get that:

$$ \sin \theta = \pm \frac{1}{\sqrt{2} }$$

I'm having a hard time interpreting the above result, I understand that when I apply the partial derivative condition and plug that back into time independent projectile motion equation, I get an equation which gives me the peak of a parabola from the family of parabolas but I can't get these points:

1. Why is it that this is also the condition that I throw the projectile such that it reaches the intended point in least speed? Like how do I get the intuition behind this?
2. When I plug $y=0$ in (1), I am finding the x intercept of parabola, but if I recall correctly, the envelope goes over the peaks of all the possible parabolas, so am I finding a parabola who's peak is the x intercept? Am I misunderstanding something?

Answer 1

1. Imagine a ball was launched at (A) $45^\circ$ to the horizontal, it has the maximum range (i.e maximum x when y=0). The ball launched at $60^\circ$ (B) at the same speed, has a smaller range.

You can see this with a quick sketch.

It also means that if you wanted the ball to reach the same place it landed at in (A), but with an angle of $60^\circ$, you'd need a higher speed. So the $45^\circ$ angle also gives you the minimum speed to reach a given point on the ground.

2. Yes you're finding the maximum x intercept.

In your formula

$0 = x \tan \theta - \frac{g}{2} ( \frac{x}{u})^2 \sec^2 \theta$

it will find the two x intercepts by factorising like this

$x (\tan \theta - \frac{g}{2} \frac{x}{u^2} \sec^2 \theta) = 0$

one solution is $x=0$, the starting point, or using

$sec^2 \theta = 1+tan^2\theta $

the other x value is

$$X=A\frac{t}{1+t^2}\tag{*}$$

where A is the constant $\frac{2u^2}{g}$, and $t=tan\theta$. You can then ignore the A and differentiate to find the best $t$ for the maximum range, $X$. That gives $t^2=1$ and the angle is $45^\circ$.

Alternatively put $t=\frac{sin\theta}{cos\theta}$ in $*$, multiply top and bottom by $cos^2\theta$, to get to $Asin\theta cos\theta$ or $\frac12Asin(2\theta)$ and that also has a maximum at $45^\circ$.

(the sentence under your equation 1 seems a bit hard to imagine, you have the right answer, but people would usually think of "maximizing x, when y=0, for different angles of projection")