Envelope of Trajectories

This was a fun one. Suppose the particle is launched with speed $u$ at angle $\theta$ .

Time at which particle gets to horizontal position $x$ :

$x = u \cos \theta \, t \\ t = \frac{x}{u \cos \theta}$

Vertical position at that same time:

$y = u \sin \theta \, t - \frac{1}{2} g t^2 \\ = x \tan \theta - \frac{g x^2}{2 u^2} sec^2 \theta$

Differentiate the $y$ expression with respect to $\theta$ and set the resulting expression to zero in order to find the $\theta$ value which maximizes $y$ for a given $x$ . The result is:

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

At this point I got lazy. I swept x from $0$ to $40$ , with $40$ being the horizontal range for $\theta = \pi/4$ . For each $x$ , I calculated the optimal $\theta$ and the resulting maximum $y$ .

$\int_0^{40} = y_{max} (x) \, dx = \frac{1600}{3}$

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import math

dx = 40.0/(10.0**7.0)

u = 20.0
g = 10.0

usq = u**2.0

#######################

A = 0.0

x = dx

while x <= 40.0:

    theta = math.atan(usq/(g*x))

    v0y = u*math.sin(theta)
    t = x/(u*math.cos(theta))

    y = v0y*t - 0.5*g*(t**2.0)

    #print x,y

    dA = y*dx
    A = A + dA

    x = x + dx

#######################

print ""
print ""
print dx
print A
print (3.0*A)

#4e-06
#533.333293313
#1599.99987994
#>>> 

Begin by forming the the particles vertical position $y$ as a function of its horizontal position $x$ from the kinematic relationships:

$y = v \sin (\theta) t - \frac{1}{2} g t^2 \quad \quad \text{ Eq1 }$

$x = v \cos (\theta) t \quad \quad \text{ Eq2 }$

Eliminate the time parameter:

$y = \tan (\theta) x - \frac{g}{2 v^2 \cos^2 (\theta) } x^2 \quad \quad \text{ Eq3 }$

$x = \frac{v^2}{g} \sin ( 2\theta ) \quad \quad \text{ Eq4 }$

At any given value of horizontal position $x$ , we want to find the launch angle $\theta$ that maximizes the vertical height $y$ . We can treat $x$ as a parameter and differentiate $\text{ Eq3 }$ w.r.t $\theta$ .

$\frac{dy}{d \theta} = \frac{1}{\cos^2 ( \theta ) }x - \frac{g \sin ( \theta ) }{v^2 \cos^3 ( \theta ) } x^2 \quad \quad \text{ Eq5 }$

Setting the result equal to $0$ we find that the tangent of the angle that maximizes the vertical height for a given $x$ to be:

$\tan ( \theta ) = \frac{v^2}{g x} \quad \quad \text{ Eq6 }$

Using Trigonometric Identity $\tan^2 ( \theta ) + 1 = \frac{ 1 }{ \cos^2 (\theta) }$ we find that:

$\frac{ 1 }{ \cos^2 ( \theta) } = 1 + \left( \frac{v^2}{g x} \right)^2 \quad \quad \text{ Eq7 }$

Now we can substitute $\text{ Eq6 \& Eq7 }$ into $\text{ Eq3 }$ . This will be the equation of the "envelope" $Y(x)$ :

$Y(x) = \frac{v^2}{2 g} - \frac{g}{2 v^2} x^2 \quad \quad \text{ Eq8 }$

Now, we are asked for the area bound by the "envelope" $Y(x)$ . We are left to determine the limits of integration by finding the maximum possible range $x$ of the projectile. From $\text{ Eq4 }$ we can differentiate w.r.t $\theta$ .

$\frac{dx}{d \theta} = \frac{2 v^2}{g} \cos ( 2\theta ) \quad \quad \text{ Eq9 }$

Equating $\text{ Eq9 }$ to $0$ we find:

$\begin{aligned} \cos ( 2 \theta ) &= 0 \\ & \Updownarrow \\ 2 \theta &= \frac{ \pi }{2} \\ \theta &= \frac{ \pi }{4} \end{aligned}$

(This is a a well known result )

From $\text{Eq4}$ we have that:

$\begin{aligned} x_{max} &= \frac{v^2}{g} \sin ( 2 \frac{ \pi }{4}) \\ &= \frac{v^2}{g} \end{aligned}$

Finally we integrate $Y(x)$ from $0 \to x_{max}$ and plug in known values:

$\displaystyle \begin{aligned} \int\limits_0^{x_{max}} Y(x) dx &= \frac{v^2}{2 g}x - \frac{g}{6 v^2} x^3 \bigg|_0^{ x_{max}} \\ \quad \\ &= \frac{v^2}{2 g}x_{max} - \frac{g}{6 v^2} x_{max}^3 \\ \quad \\ &= \frac{1}{3} \frac{v^4}{g^2} \\ \quad \\ &= \SI{\frac{1600}{3}}{m^2} \end{aligned}$

Good problem! I had to think about this for a while. First time actually calculating an "envelope" I believe.