Volume 3: The Calculus of Life

Workbook 28.3: Projectile Motion

Directives for the Master Archer:

1. X-Progress: $x(t) = (v_0 \cos \theta)t$. (Horizontal).
2. Y-Progress: $y(t) = -16t^2 + (v_0 \sin \theta)t + h_0$. (Vertical).
3. The Peak: Happens when $y'(t) = 0$. Solve for $t$, then find $y$.
4. The Range: Happens when $y(t) = 0$. Solve for $t$, then find $x$.

Part I: The Launch Setup

An arrow is launched from the ground ($h_0=0$) with $v_0 = 120$ ft/s at an angle of $45^\circ$.
Note: $\cos(45^\circ) = \sin(45^\circ) = 0.707$.

The Components: Find the horizontal speed ($v_x$) and the initial vertical speed ($v_{y0}$).

$v_x = 120 \cdot 0.707 = ...$
$v_{y0} = 120 \cdot 0.707 = ...$

The Peak Time: At what time $t$ does the arrow reach its highest point?

$y'(t) = -32t + v_{y0} = 0$.
Solve for $t$: ...

The Maximum Height: Plug the time from above into the $y(t)$ equation.

$y(t) = -16t^2 + v_{y0} \cdot t = ...$

Part II: The Impact (Range)

Total Flight Time: For how many seconds does the arrow stay in the air? (Hint: $2 \times t_{\text{peak}}$).

...

Total Range: How far away from the archer does the arrow hit the ground? (Use your flight time in the $x(t)$ equation).

...
The Logic Check:

If you launch the arrow at 60 degrees (higher), will the Range be larger or smaller than the 45-degree launch? Will the Peak Height be larger or smaller? Why can't we have both at the same time?

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Part III: The High Ground

Nehemiah's Defense: An archer stands on top of a wall ($h_0 = 20$ feet). He shoots an arrow horizontally ($0^\circ$ angle) at 100 ft/s.
How long until the arrow hits the ground ($y=0$)? How far from the wall will it land?

$y(t) = -16t^2 + 0 \cdot t + 20 = 0$.
Solve for $t$... then find $x(t)$...

Part IV: The Challenge (The Mark)

The Giant Slayer

A slinger launches a stone at a giant 200 feet away. The stone must hit the giant at a height of exactly 10 feet. If the stone is launched at 100 ft/s at an angle of $30^\circ$...
Task: Will the stone hit the mark? Find the height of the stone when its horizontal distance ($x$) is exactly 200 feet.

Step 1: Solve $x(t) = 200$ to find the time...
Step 2: Plug that time into $y(t)$...
Result: ...

Part V: Transmission (The Echad Extension)

Teacher Log: The Arc of Love

Objective: Explain Projectile Motion to a younger student using a hose or a ball.

The Activity:
1. Throw a ball in a high arc.
2. Throw it in a flat line.
3. Ask: "Which way looks more like an arrow?"

The Lesson: "Our life is like a throw. We have to aim high to beat the gravity of the world and land exactly where God wants us."


Response: ___________________________________________________________

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