Constrained Projectile Motion Problems

Hello! Welcome to our next lesson on advanced projectile motion.

In our previous session, we derived and applied the equation for the envelope of trajectories, which defines the absolute boundary of all reachable points for a given launch speed. We focused on the question: can a projectile hit a certain region?

Today, we move from possibility to specifics. Our learning outcome is to solve complex projectile problems involving constraints, such as hitting a target at a specific time and location. We will investigate the conditions required to hit a precise point $(X, Y)$ and see how adding a time constraint leads to a unique solution. This involves finding the necessary launch parameters—speed $v_{0}$ and angle $θ$ —that satisfy these constraints.

1. Hitting a Target: Finding the Required Launch Angle

Let's start with the most common scenario: you know your launch speed $v_{0}$ , and you have a target at coordinates $(X, Y)$ . The question is, at what angle $θ$ should you launch?

Your strong mathematical background will be very useful here. The trajectory equation, which we've used before, can be rearranged into a quadratic equation for $tan θ$ . The roots of this equation give us the required launch angle(s).

To see the full derivation, please study the following resource.

Contents

The 'Modelling Motion' workbook provides a rigorous derivation for this exact problem. It elegantly transforms the trajectory equation into a quadratic form that we can analyze.

Please read the part of the document titled 'Choosing trajectories,' starting from the beginning of that subsection and continuing up to the start of 'Example 5'. Pay close attention to the derivation of Equation (1.14), which is the quadratic equation in terms of tan(θ). It's crucial to understand why the condition for real roots (Equation 1.15) determines if the target is reachable.

As you've just read, for a given launch speed $v_{0}$ and a target $(X, Y)$ , the trajectory equation:
$Y = X tan θ - \frac{g X ^{2}}{2 v _{0}^{2}} (1 + tan^{2} θ)$
can be rearranged into the quadratic form $A (tan θ)^{2} + B (tan θ) + C = 0$ , where:
$A = \frac{g X ^{2}}{2 v _{0}^{2}}, B = - X, C = Y + \frac{g X ^{2}}{2 v _{0}^{2}}$

The solutions to this quadratic, given by $tan θ = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ , reveal the physics:

Two real roots: If the discriminant $B^{2} - 4 A C > 0$ , there are two distinct launch angles that will hit the target. This corresponds to a point inside the parabola of safety. One angle gives a low, direct trajectory, and the other a high, arcing trajectory.
One real root: If $B^{2} - 4 A C = 0$ , there is exactly one launch angle. This corresponds to a target located precisely on the parabola of safety. You must launch at the one angle that allows you to just reach it.
No real roots: If $B^{2} - 4 A C < 0$ , the target is unreachable with the given launch speed. It lies outside the parabola of safety.

This analysis provides a deep connection between the algebraic nature of the solution and the physical boundary we studied in the last lesson.

Example: The Tiddly-winks Problem

Let's look at Example 6 (choosing angle) from the resource you just read. Here, a player wants a "wink" to land in a cup. The constraints are:

Launch speed, $u = 1.5$ m/s.
Target coordinates relative to launch, $(x, y) = (0.1 m, 0.05 m)$ .

Plugging these into the quadratic equation for $tan θ$ yields two solutions: $θ = 41. 5^{\circ}$ and $θ = 71. 8^{\circ}$ . As the example explains, the steeper angle is preferred because the wink is descending as it reaches the cup, making it more likely to be "potted".

2. The Time-Constrained Problem: A Unique Solution

Now, let's add the ultimate constraint: time. What if we must not only hit the target $(X, Y)$ but also arrive at a precise time $T$ ? This is akin to planning a rendezvous or an interception.

This constraint fundamentally changes the problem. Instead of multiple possible trajectories, we will find there is only one unique launch velocity vector that satisfies the conditions. The most direct way to prove this is by using the parametric equations of motion.

Let's derive this from first principles. Our goal is to find the initial velocity components, $v_{0 x}$ and $v_{0 y}$ , given the target $(X, Y)$ and the time of flight $T$ . The origin is the launch point $(0, 0)$ .

The parametric equations are:

$x (t) = v_{0 x} t$
$y (t) = v_{0 y} t - \frac{1}{2} g t^{2}$

At time $t = T$ , the projectile must be at $(X, Y)$ . Substituting these values in:

$X = v_{0 x} T$
$Y = v_{0 y} T - \frac{1}{2} g T^{2}$

This gives us a simple system of two linear equations for the two unknowns, $v_{0 x}$ and $v_{0 y}$ .

From equation (1), we solve for the horizontal velocity component:
$v_{0 x} = \frac{X}{T}$

From equation (2), we solve for the vertical velocity component:
$v_{0 y} T = Y + \frac{1}{2} g T^{2} ⟹ v_{0 y} = \frac{Y}{T} + \frac{1}{2} g T$

As you can see, both $v_{0 x}$ and $v_{0 y}$ are uniquely determined. There is no ambiguity. Once you specify a target and a time of flight, there is only one possible launch velocity vector to achieve it.

From these components, we can find the launch speed $v_{0}$ and angle $θ$ :

$v_{0} = v_{0 x}^{2} + v_{0 y}^{2}$
$θ = arctan (\frac{v _{0 y}}{v _{0 x}})$

This rigorous approach is also explored in the context of a programming problem in the following resource.

Find projectile speed given maximum height and range.

This derivation is so fundamental that it appears in many contexts, including computer game physics. The following answer on Math StackExchange formalizes the exact derivation we just performed.

Please read the answer by user Xander Henderson. Focus on how the problem is set up using the parametric equations for the position q(t). Observe how specifying the time-to-target, denoted as t_t, leads directly to unique expressions for the initial horizontal velocity (a) and the initial vertical velocity (b).

Test your understanding!

A fireworks shell is launched from the ground. It must explode at a height of 180 m and a horizontal distance of 120 m from the launch site. For the best visual effect, the explosion must happen exactly 6 seconds after launch. What is the required initial launch velocity (magnitude and angle)? Use $g \approx 10 m/s^{2}$ .

Show answer

We are given:

Target coordinates $(X, Y) = (120 m, 180 m)$ .
Time of flight $T = 6 s$ .

First, find the required initial velocity components:

$v_{0 x} = \frac{X}{T} = \frac{120 m}{6 s} = 20 m/s$ .
$v_{0 y} = \frac{Y}{T} + \frac{1}{2} g T = \frac{180 m}{6 s} + \frac{1}{2} (10 m/s^{2}) (6 s) = 30 m/s + 30 m/s = 60 m/s$ .

Now, calculate the magnitude and angle of the launch velocity:

Magnitude: $v_{0} = v_{0 x}^{2} + v_{0 y}^{2} = (20)^{2} + (60)^{2} = 400 + 3600 = 4000 = 2010 \approx 63.2 m/s$ .
Angle: $θ = arctan (\frac{v _{0 y}}{v _{0 x}}) = arctan (\frac{60}{20}) = arctan (3) \approx 71. 6^{\circ}$ .

The shell must be launched with a speed of approximately 63.2 m/s at an angle of 71.6° above the horizontal.

3. Advanced Application: Optimization with Constraints

To conclude, let's look at a problem that synthesizes these ideas with the calculus of optimization. What if we want to maximize range, but we are not launching from ground level? This is a classic, advanced problem that requires careful application of the trajectory equation and differentiation.

Example 7 (choosing the angle again) from the HELM (2008) workbook provides a fantastic analysis of this. A shot-putter launches from a height $H$ and wants to find the optimal angle $θ$ to maximize the horizontal range $R$ . The derivation uses implicit differentiation, a technique you should find interesting, to arrive at the condition for the optimal angle. I encourage you to work through this example in the resource on your own to challenge your understanding. It shows that for a launch from an elevation ( $H > 0$ ), the optimal angle for maximum range is always less than 45°.

Conclusion

Today, we've moved from the general to the specific, developing rigorous methods to solve for the exact launch conditions needed to hit a constrained target.

Key Takeaways:

To hit a target $(X, Y)$ with a given speed $v_{0}$ , we solve a quadratic equation for $tan θ$ . If the target is reachable, there are typically two possible launch angles.
To hit a target $(X, Y)$ at a specific time $T$ , the problem has a unique solution. The initial velocity components $v_{0 x}$ and $v_{0 y}$ are completely determined by the constraints.
Adding constraints to a kinematics problem reduces the number of possible solutions. A spatial constraint allows for multiple trajectories, while a spatio-temporal constraint defines a single, unique trajectory.

Preview of the Next Lesson:
We have now concluded our deep dive into advanced projectile motion. Our next module marks a shift in the type of motion we analyze. We will move from trajectories in a plane to motion along a circular path. In the first lesson of Module 6, we will define angular position, velocity, and acceleration, and derive their relationships using calculus. This will lay the groundwork for understanding the kinematics of rotation, a crucial topic in mechanics.

Recommended resources

1. Hitting a Target: Finding the Required Launch Angle

2. The Time-Constrained Problem: A Unique Solution

3. Advanced Application: Optimization with Constraints

Conclusion