Can You Draft A Riddler Fantasy Football Dream Team?

A runner catches the ball on one goal line and sprints at $15$ mph straight down the field, $100$ yards to the far end zone. You start on the same goal line, $50$ yards to his side, and always run directly at his current position (a pure pursuit, never leading him). How fast must you run to catch him before he scores?

Solution

Place yourself at the origin and the runner at $(x_0, 0)$ with $x_0 = 50$, running up the line $x = x_0$ at speed $V_r$. You move at speed $V_c$ always aimed at his current spot, tracing a pursuit curve $y(x)$. Two facts pin the curve down: your heading points at the runner, so the tangent satisfies $\frac{dy}{dx} = \frac{V_r t - y}{x_0 - x}$, and the arc length you have covered is $V_c t = \int_0^x \sqrt{1 + (y')^2}\,dz$. Eliminating $t$ and differentiating once turns this into a separable equation for the slope $p = y'$, with $n = V_r/V_c$: $$(x_0 - x)\,\frac{dp}{dx} = n\sqrt{1 + p^2}.$$ Integrating from the start (where $p = 0$ at $x = 0$) gives $\big(p + \sqrt{1 + p^2}\big)\big(1 - \tfrac{x}{x_0}\big)^{n} = 1$, and integrating once more yields the capture height: the runner is overtaken on his line $x = x_0$ at $$y_{\text{catch}} = \frac{n}{1 - n^2}\,x_0 \qquad (n < 1).$$ You catch him before the end zone exactly when $y_{\text{catch}} \le 100$; the slowest speed that just manages it sets equality: $$100 = \frac{n}{1 - n^2}\cdot 50 \;\Longrightarrow\; 2n^2 + n - 2 = 0 \;\Longrightarrow\; n = \frac{\sqrt{17} - 1}{4}.$$ Since $V_c = V_r/n$, $$V_c = \frac{15}{n} = \frac{\sqrt{17} + 1}{4}\cdot 15 \approx \boxed{19.2 \text{ mph}}.$$

The computation

Run the chase directly: step the runner up his line, step yourself the same time-step straight toward him, and binary-search the smallest speed that closes the gap before he travels $100$ yards.

import numpy as np
Vr, x0, goal = 15.0, 50.0, 100.0
def caught(Vc, dt=1e-5):
    cx = cy = ry = 0.0
    while ry < goal:
        ry += Vr * dt
        dx, dy = x0 - cx, ry - cy
        d = np.hypot(dx, dy)
        if d < 5e-3: return True
        cx += Vc * dt * dx / d; cy += Vc * dt * dy / d
    return False
lo, hi = 15.0, 30.0
for _ in range(40):
    mid = (lo + hi) / 2
    hi, lo = (mid, lo) if caught(mid) else (hi, mid)
print(hi, 15 * (np.sqrt(17) + 1) / 4)         # ~19.2, 19.21