Determining the initial angle of a projectile's motion

Question Q8.4.3

The trajectory of a projectile in the $xz$ -plane launched from the origin at an angle $\theta_0$ from the (horizontal) $x$ -axis with speed $v_0 = 25\;\mathrm{m\,s^{-1}}$ is $z = x\tan\theta_0 - \frac{g}{2v_0^2\cos^2\theta_0}x^2.$

Derivation

During its flight, the projectile is subject to a constant acceleration

$\ddot{\boldsymbol{r}} = -g\hat{\boldsymbol{k}}$ . Integrating,

$\begin{align*} \boldsymbol{v} &= \dot{\boldsymbol{r}} = -gt\hat{\boldsymbol{k}} + \boldsymbol{v_0}\\ \boldsymbol{r} & = \textstyle -\frac{1}{2}gt^2\hat{\boldsymbol{k}} + \boldsymbol{v_0}t + \boldsymbol{r_0} \end{align*}$ where

$\boldsymbol{v_0} = (v_0\cos\theta)\hat{\boldsymbol{i}} + v_0(\sin\theta)\hat{\boldsymbol{k}}$ and

$\boldsymbol{r_0}=0$ (the projectile starts at the origin). Therefore,

$x = v_0t\cos\theta$ and

$z = v_0t\sin\theta - \frac{1}{2}gt^2$ . Eliminating

$t$ from these expressions gives the formula for

$z$ above.

If the projectile passes through the point $(5, 15)$ , use Brent's method to determine the possible values of $\theta_0$ .

Solution Q8.4.3

In general, there are two (physically distinct) possible angles $\theta_0$ corresponding to the projectile passing through the specified point, $(x_1, y_1) = (5,15)$ , on the way up or on the way down. These values are the roots in $(0, \pi/2)$ of the function $f(\theta_0; x_1, z_1) = x_1\tan\theta_0 - \frac{g x_1^2}{2v_0^2\cos^2\theta_0} - z_1$ After bracketing the roots with a rough plot of $f(\theta_0)$ , we can use brentq:

In [x]: g = 9.81
In [x]: v0, x1, z1 = 25, 5, 15
In [x]: f = lambda theta0, x1, z1: x1 * np.tan(theta0) - g / 2\
                                   * (x1 / v0 / np.cos(theta0))**2 - z1
In [x]: th1 = brentq(f, 1, 1.4, args=(x1,z1))
In [x]: th2 = brentq(f, 1.5, 1.6, args=(x1,z1))
In [x]: np.degrees(th1), np.degrees(th2)
Out[x]: (74.172740936822834, 87.392310240255171)

That is, $\theta_0 = 74.2^\circ$ or $\theta_0 = 87.4^\circ$ .