Ballistic problem – Maximum horizontal reach when firing toward a high place

Here is an example of the application of Newton’s second law of motion to a ballistic problem. The problem we want to consider is calculating the critical angle \(\theta_c\) of firing a projectile from the ground toward a high place (of height \(h\)), located at a distance \(L\) in front of the place of firing, in order to reach the maximum horizontal distance \(z_{\text{max}}\) on the high place. It is supposed that air resistance is negligible and that the initial velocity \(v_0\) is a known value. The problem is easy to solve however nontrivial in the sense that there could be no solution. Then, part of the problem is to find the admissible values of the parameters \(L\), \(h\), and \(v_0\) such that a solution exists.

Ballistic problem

Let’s start by recalling the equations of motion in two dimensions:

\begin{equation}
\begin{split}
m\frac{d^2x}{dt^2}&=0\\
m\frac{d^2y}{dt^2}&=-mg ,\\
\end{split}
\end{equation}

whose respective solutions are

\begin{equation}
\begin{split}
x(t)&=x_0+(v_0\cos\theta)t\\
y(t)&=y_0+(v_0\sin\theta)t-\frac{1}{2}gt^2
\end{split}
\end{equation}

In the present case it is convenient to select the origin of coordinates at the place of firing so that \((x_0,y_0)=(0,0)\). Then, we can put \(t=x/(v_0\cos\theta)\) on the equation for \(y\) to get the following relation between the coordinates:

\begin{equation}
y=(\tan\theta)x-\frac{g}{2v_0^2\cos^2\theta}x^2.
\end{equation}

Equation (3) is the starting point to solve the problem.

Ballistic problem

Maximum horizontal reach on the high terrain

When the projectile reaches the high terrain, the \(y\) coordinate takes the value \(h\), and the coordinate \(x\) takes the value \(L+z\). So, equation (3) becomes

\begin{equation}
h=(\tan\theta)(L+z)-\frac{g}{2v_0^2\cos^2\theta}(L+z)^2.
\end{equation}

We are looking for the angle \(\theta_c\) such that \(z\) reaches its maximum possible value. This means that \(\theta_c\) is determined by the condition \(dz/d\theta=0\). Implicit differentiation of equation (4) leads to

\begin{equation}
\begin{split}
&\left(\frac{1}{\cos^2\theta}\right) (L+z)+(\tan\theta)\frac{dz}{d\theta}\\
&-\frac{g\sin\theta}{v_0^2\cos^3\theta}(L+z)^2-\frac{g}{2v_0^2\cos^2\theta}\frac{dz}{d\theta}=0.
\end{split}
\end{equation}

Now we apply the condition \(dz/d\theta=0\) when \(\theta=\theta_c\):

\begin{equation}
\left(\frac{1}{\cos^2\theta_c}\right)(L+z_{\text{max}})-\frac{g\sin\theta_c}{v_0^2\cos^3\theta_c}(L+z_{\text{max}})^2=0.
\end{equation}

From equation (6) we obtain the relation

\begin{equation}
L+z_{\text{max}}=\frac{v_0^2}{g\tan\theta_c}.
\end{equation}

Now we plug that expression in equation (4):

\begin{equation}
h=\frac{v_0^2}{g}-\frac{v_0^2}{2g\tan^2\theta_c\cos^2\theta_c}
\end{equation}

The latter equation is readily solved to get

\begin{equation}
\boxed{\sin\theta_c=\frac{1}{\sqrt{2(1-gh/v_0^2)}}}.
\end{equation}

The latter equation imposes a restriction on the possible values the parameters \(h\) and \(v_0\) can take, namely \(\frac{2gh}{v_0^2}\leq1\). Other wise, the critical angle is not defined (remember that \(\sin\theta\leq1\)).

The maximum horizontal reach is obtained by plugin the latter result in equation (7) and using the relation \(\cos\theta=\sqrt{1-\sin^2\theta}\). The result is

\begin{equation}
\boxed{z_{\text{max}}=\frac{v_0^2}{g}\sqrt{1-2gh/v_0^2}-L}
\end{equation}

The latter equation imposes an additional restriction on the possible values of the parameters. Indeed, the condition \(z_{\text{max}}\geq0\), together with \(\frac{2gh}{v_0^2}\leq1\) and equation (10) implies

\begin{equation}
\boxed{\frac{2gh}{v_0^2}+\left(\frac{gL}{v_0^2}\right)^2\leq 1}
\end{equation}

Ballistic problem. Parameter space
The parabolic region in blue corresponds to the cases when the projectile can reach the high terrain. The vertical axis corresponds to the ratio between the potential \(mgh\) and kinetic \(mv_0^2/2\) energy. The horizontal axis corresponds to the ratio between the distance \(L\) and the maximum horizontal reach \(v_0^2/g\) when the places of firing and landing are at the same height \(h=0\).

Let’s interpret the last result. If \(L=0\), then the projectile can be fired vertically upward only. Its kinetic energy \(mv_0^2/2\) allows it to reach a maximum height given by conservation of energy. i.e., \(mv_0^2/2=mgh\). In this case, it is clear that the horizontal reach on the high terrain is zero. On the other side, if \(h=0\), then the land is flat and the maximum admissible value for \(L\) is just the maximum horizontal reach \(v_0^2/g\).

Ballistic problem. Zmax is zero
When \(L=0\) the projectile has to be fired vertically upward. The minimum kinetic energy \(mv_0^2/2\) necessary for reaching the high terrain has to be equal to the potential energy \(mgh\) at that height.

In addition, any possible combination of \(L\), \(h\), and \(v_0\) that makes the left side of equation (11) exactly equal to one, corresponds to the case when the projectile just reaches the high place, and the horizontal distance on it is zero.

ballistic problem
Any combination of \(h\), \(L\), and \(v_0\) that satisfy the equality \((2gh/v_0^2)+(gL/v_0^2)^2=1\) corresponds to the cases when the projectile just hit the corner of the high terrain.

Any other combination of the parameters that fulfill the strict inequality in equation (11) corresponds to \(z_{\text{max}}>0\). Finally, any combination of the parameters that make the left side of equation (11) greater than one, corresponds to cases where it is impossible to reach the high terrain.

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