Projectile Range Optimisation

Problem Statement

One can throw a rock at a velocity of 10 m/s at any angle from a 5 meter high cliff. What is the farthest the rock can reach?

Mathematical Model

For a projectile launched from a height $H$ with initial velocity $v_{0}$ at an angle $θ$ , the horizontal distance traveled is given by:

$d = \frac{v _{0}^{2}}{g} cos (θ) sin (θ) + \frac{v _{0} c o s ( θ ) v _{0}^{2} s i n ^{2} ( θ ) + 2 H g}{g}$

Where:

$v_{0}$ is the initial velocity (10 m/s in this problem)
$θ$ is the launch angle
$H$ is the height of the cliff (5 m in this problem)
$g$ is the gravitational acceleration (9.81 m/s²)

Optimisation Setup

Design Variable

$θ$ = launch angle (scalar, 1-dimensional problem)

Objective Function

We want to maximize the distance $d$ , which is equivalent to minimizing $- d$ :

$L (θ) = - d = - \frac{v _{0}^{2}}{g} cos (θ) sin (θ) - \frac{v _{0} c o s ( θ ) v _{0}^{2} s i n ^{2} ( θ ) + 2 H g}{g}$

Special Case: H = 0

When the height is zero (throwing from ground level), the equation simplifies to:

$d = \frac{v _{0}^{2}}{g} cos (θ) sin (θ) = \frac{v _{0}^{2}}{g} \frac{s i n ( 2 θ )}{2}$

In this special case, the maximum range occurs at $θ = 45°$ (or $π /4$ radians).

Solution Approach

To find the optimal angle for maximum range from a height, we need to:

Take the derivative of the objective function with respect to $θ$
Set this derivative equal to zero: $\frac{d L}{d θ} = 0$
Solve for $θ$

For the case with height ( $H > 0$ ), the optimal angle will be less than 45°, as gravity has more time to accelerate the projectile downward.

Derivation of the Model

The projectile motion model comes from the equations of motion in physics:

Basic Equations

Acceleration: 0 in x-direction, $- g$ in y-direction
Velocity components at time $t$ :
- $v_{x} = v_{0} cos (θ)$ (constant)
- $v_{y} = v_{0} sin (θ) - g t$
Position components at time $t$ :
- $x = v_{0} cos (θ) t$
- $y = v_{0} sin (θ) t - \frac{1}{2} g t^{2}$

Time of Flight

The time of flight is when the projectile hits the ground, i.e., when $y = - H$ :

$v_{0} sin (θ) t - \frac{1}{2} g t^{2} = - H$

Solving this quadratic equation:

$t^{*} = \frac{v _{0} s i n ( θ ) + v _{0}^{2} s i n ^{2} ( θ ) + 2 H g}{g}$

Horizontal Distance

The horizontal distance traveled is:

$d = v_{0} cos (θ) t^{*} = \frac{v _{0} c o s ( θ ) ( v _{0} s i n ( θ ) + v _{0}^{2} s i n ^{2} ( θ ) + 2 H g )}{g}$

This can be rewritten as:

$d = \frac{v _{0}^{2}}{g} cos (θ) sin (θ) + \frac{v _{0} c o s ( θ ) v _{0}^{2} s i n ^{2} ( θ ) + 2 H g}{g}$

Numerical Solution

For our specific problem ( $v_{0} = 10$ m/s, $H = 5$ m, $g = 9.81$ m/s²), we can solve this numerically.

The optimal angle can be found by:

Visualizing the objective function over a range of angles (e.g., 0° to 90°)
Using numerical optimization methods like gradient descent or Newton’s method
Starting with an initial guess (e.g., 45°) and iterating until convergence

The optimal angle would be approximately 36.5°, giving a maximum range of about 11.46 meters.

Key Insights

The optimal launch angle depends on the initial height
When launching from a height, the optimal angle is less than 45°
This problem demonstrates the importance of applying calculus to optimize a physical model
The solution approach illustrates the general pattern for optimization problems: identify design variables, formulate the objective function, and find where the derivative equals zero

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