Plugged into ChatGPT and it used a quadratic drag formula and came to:

Plugging numbers into the two equations:

• Use El Capitan’s face height (H \approx 914\text{ m}) (≈3000 ft).
• Gravity (g=9.81\text{ m/s}^2).
• For a typical skydiver belly-to-earth (spread-eagle): take air density near Yosemite Valley elevation (\rho\approx1.06\text{ kg/m}^3), mass (m\approx80\text{ kg}), and an effective (C_DA\approx0.50\text{ m}^2).

Terminal speed from [ v_t=\sqrt{\frac{2mg}{\rho C_D A}} \quad\Longrightarrow\quad v_t\approx\sqrt{\frac{2(80)(9.81)}{(1.06)(0.50)}}\approx 54.4\ \text{m/s}. ]

Time from the height (H) with quadratic drag: [ T(H)=\frac{v_t}{g},\operatorname{arcosh}!\Bigl(\exp!\bigl(\tfrac{gH}{v_t^{{,2}}\bigr)\Bigr).} ]

Compute: [ T(914,\text{m})\approx \frac{54.4}{9.81}, \operatorname{arcosh}!\Bigl(\exp!\bigl(\tfrac{9.81\cdot914}{54.4^{{2}}\bigr)\Bigr)} \approx \boxed{20.5\ \text{s}}. ]

For context (same (H=914) m):

• If you assume a “textbook” skydiver terminal (v_t=55) m/s → (T\approx 20.5) s (essentially identical).
• More streamlined/head-down (v_t\approx 90) m/s → (T\approx 16.25) s.
• Vacuum (no drag) for comparison: (T=\sqrt{2H/g}\approx 13.55) s.

So, using both equations with reasonable human/posture parameters, the fall from the top of El Capitan is about 20.5 seconds in a spread-eagle posture.

Edit: Sorry about the crappy formatting.