Derivation: Focal Distance Property
Proof: Equation of the Directrix
We must show that for a projectile with initial speed $v_0$, the directrix is always at height $H = \frac{v_0^2}{2g}$.
1. The Trajectory Equation
A projectile launched at angle $\theta$ follows the path: $$ y = x \tan\theta – \frac{g x^2}{2 v_0^2 \cos^2\theta} $$
A projectile launched at angle $\theta$ follows the path: $$ y = x \tan\theta – \frac{g x^2}{2 v_0^2 \cos^2\theta} $$
2. Converting to Vertex Form
Completing the square for $x$: $$ \left( x – \frac{v_0^2 \sin\theta \cos\theta}{g} \right)^2 = -\frac{2 v_0^2 \cos^2\theta}{g} \left( y – \frac{v_0^2 \sin^2\theta}{2g} \right) $$
Completing the square for $x$: $$ \left( x – \frac{v_0^2 \sin\theta \cos\theta}{g} \right)^2 = -\frac{2 v_0^2 \cos^2\theta}{g} \left( y – \frac{v_0^2 \sin^2\theta}{2g} \right) $$
3. Calculating Directrix Height
For a downward-opening parabola with Vertex $k$ and Focal Length $a$:
For a downward-opening parabola with Vertex $k$ and Focal Length $a$:
- Vertex Height: $k = \frac{v_0^2 \sin^2\theta}{2g}$
- Focal Length ($4a = \text{coeff}$): $a = \frac{v_0^2 \cos^2\theta}{2g}$
Application to the Problem
Using the property derived above ($H = v_0^2/2g$), we established that minimizing $v_0$ is equivalent to minimizing $H$. Now, we derive the exact value for this minimum height.
Derivation of $H_{min}$
We want to find the lowest possible height $H$ for the directrix such that a parabola with focus $F$ can still pass through points $P_1$ and $P_2$.
1. Expressing Distances via Focus
By the definition of a parabola, the distance from any point on the curve to the Focus ($F$) equals its vertical distance to the Directrix ($y=H$). $$ |FP_1| = H – h_1 $$ $$ |FP_2| = H – h_2 $$
By the definition of a parabola, the distance from any point on the curve to the Focus ($F$) equals its vertical distance to the Directrix ($y=H$). $$ |FP_1| = H – h_1 $$ $$ |FP_2| = H – h_2 $$
2. Applying the Triangle Inequality
Consider the triangle formed by points $P_1$, $P_2$, and the Focus $F$. The length of any side of a triangle must be less than or equal to the sum of the other two sides.
Specifically, the straight-line distance $d$ between $P_1$ and $P_2$ satisfies: $$ |P_1 P_2| \le |FP_1| + |FP_2| $$ $$ d \le (H – h_1) + (H – h_2) $$
Consider the triangle formed by points $P_1$, $P_2$, and the Focus $F$. The length of any side of a triangle must be less than or equal to the sum of the other two sides.
Specifically, the straight-line distance $d$ between $P_1$ and $P_2$ satisfies: $$ |P_1 P_2| \le |FP_1| + |FP_2| $$ $$ d \le (H – h_1) + (H – h_2) $$
3. Solving for the Minimum H
We rearrange the inequality to isolate $H$: $$ d \le 2H – (h_1 + h_2) $$ $$ 2H \ge d + h_1 + h_2 $$ $$ H \ge \frac{d + h_1 + h_2}{2} $$
The smallest possible value for $H$ ($H_{min}$) occurs when the inequality becomes an equality. Geometrically, this happens when the triangle “collapses” into a straight line, meaning the Focus $F$ lies exactly on the segment connecting $P_1$ and $P_2$. $$ H_{min} = \frac{d + h_1 + h_2}{2} $$
We rearrange the inequality to isolate $H$: $$ d \le 2H – (h_1 + h_2) $$ $$ 2H \ge d + h_1 + h_2 $$ $$ H \ge \frac{d + h_1 + h_2}{2} $$
The smallest possible value for $H$ ($H_{min}$) occurs when the inequality becomes an equality. Geometrically, this happens when the triangle “collapses” into a straight line, meaning the Focus $F$ lies exactly on the segment connecting $P_1$ and $P_2$. $$ H_{min} = \frac{d + h_1 + h_2}{2} $$
Final Calculation
$$ v_{min} = \sqrt{2g H_{min}} = \sqrt{g(d + h_1 + h_2)} $$ Given $h_1 = 12$, $h_2 = 18$, $d = 10$, and $g=10$: $$ v_{min} = \sqrt{10(10 + 12 + 18)} = \sqrt{400} $$
Final Answer
$$ v_{min} = 20 \, \text{m/s} $$