Death defying skateboard stall

Here is a more traditional solution, without using the code.

The equation of motion is $\frac{d^2y}{dt^2} = \frac{-F_d - F_g}m = -\kappa \left(\frac{dy}{dt} + w\right)^2 -g,\ \ \ \ \ \ \kappa = \frac{\rho C_d A}{2m}.$ Switch to dimensionless units, as follows: $t^\star = \sqrt{\frac 1{g\kappa}} = \SI{3.343}{s};\ \ v^\star = \sqrt{\frac g \kappa} = \SI{33.43}{m/s};\ \ y^* = \frac 1 \kappa = \SI{11.77}{m}.$ $\tau = \frac t{t^\star};\ \ \ \beta = \frac v{v^\star};\ \ \ \lambda = \frac y{y^\star}.$ The equation of motion becomes $-\frac{d^2\lambda}{d\tau^2} = \left(\frac{d\lambda}{d\tau} + \omega\right)^2 + 1,$ where $\omega = w/v^\star = 0.8973$ is the dimensionless wind speed. Setting $\beta_{rel} = d\lambda/d\tau + \omega$ , this becomes $-\frac{d\beta_{rel}}{d\tau} = \beta_{rel}^2 + 1.$ Integration yields $\frac{d\lambda}{d\tau} = \beta_{rel} - \omega = \tan(\tau_0 - \tau) - \omega.$ Here, $\tau_0$ is an integration constant. It corresponds to the time when Bob will be falling down at speed $w$ .

We integrate again: $\lambda = \ln \frac{\cos (\tau_0 - \tau)}{\cos\tau_0} - \omega \tau.$ The integration constant was chosen to ensure that $\lambda = 0$ at $\tau = 0$ .

If Bob's initial speed (at $\tau = 0$ ) is $d\lambda/d\tau = \beta_0 = v_0/v^\star$ , then we find $\tan \tau_0 = \beta_0 + \omega.$ Bob reaches the highest points at $\tau = \tau_0$ , when $d\lambda/d\tau = 0$ , so that $\tan (\tau_0 - \tau_t) = \omega.$ We write for the greatest height $\lambda_t = \ln \frac{\cos\tau_0}{\cos (\tau_0 - \tau_t)} - \omega \tau_t;$ using $\cos x = 1/\sqrt{1 + \tan^2 x}$ and $\tan \tau_t = \tan(\tau_0 - (\tau_0 - \tau_t))$ , we eliminate the $\tau$ variables: $\lambda_t = \frac12 \ln \frac{1 + (\beta_0 + \omega)^2}{1 + \omega^2} - \omega\ \text{inv tan}\ \frac{\beta_0}{1 + (\beta_0 + \omega)\omega}.$ We are told that $\lambda_t = y_t/y^\star = 0.03579$ ; in principle, we could solve this equation for $\beta_0$ using numerical methods; we find $\beta_0 \approx 0.411$ and $v_0 \approx \SI{13.74}{m/s}$ .

An approximate solution may be obtained by Taylor approximations and ignoring higher powers than $\beta_0^2$ . We have $\lambda_t \approx \frac12 \left[\frac{2\omega}{1+\omega^2}\beta_0 + \frac{(1-\omega)^2}{(1+\omega^2)^2}\beta_0^2\right] - \left[\frac{\omega}{1+\omega^2}\beta_0 - \frac{\omega^2}{(1+\omega^2)^2}\beta_0^2\right] = \frac{1 - 2\omega + 3\omega^2}{(1 + \omega^2)^2}\cdot \tfrac12\beta_0^2.$ Solving, we obtain $\beta_0 \approx \frac{1 + \omega^2}{\sqrt{1 - 2 \omega + 3 \omega^2}}\sqrt{2 \lambda_t}.$ (Note: $\beta_0 = \sqrt{2 \lambda_t}$ would be the approximation without wind and ignoring air resistance, corresponding to the familiar $v_0 = \sqrt{2gh}$ .)

In our situation, this becomes $\beta_0 \approx 0.3793,\ \ \ \ v_0 \approx \SI{12.7}{m/s}.$ The fact that $\beta$ is not significantly less than unity shows that this approximation is not great. Indeed, we are 7% low-- but it gives at least an idea of the answer!

The question is written so that one might attempt to set a finite starting velocity and adjust it until velocity is zero at 4m. The other way around is much simpler and straight forward, assuming v=0 at t=0, working backward in time to what the velocity must be in order to fulfill that.

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g = 10
w = 30
m = 90
c = 0.5 * 1.22 * 1.32

t = 0
h = 0
d = 0.001 # significantly small for high resolution
v = 0 # <- don't bother experimenting with this initial value.

(times, heights) = ([], [])

while h<4:
    t+=d
    a=g + (w+v)*(w+v)*c/m
    h+=v*d + 0.5*d*d*a
    v+=d*a
    times.append(t)
    heights.append(h)

print(v) # The answer you seek
print(h) # just for kicks
print(t) # how long it took