## Fundamentals of Physics Extended (10th Edition)

$H=292.75\mbox{m}$
Height, at which parachute is opened, is given by equation $$H_{dec}=v_{max}t_{dec}-\frac{1}{2}a_{dec}t_{dec}^2=\sqrt{2gH_{freely}}\times\frac{\sqrt{2gH_{freely}}-v_{final}}{a_{decel}}-\frac{1}{2}a_{decel}\times\bigg(\frac{\sqrt{2gH_{freely}}-v_{final}}{a_{decel}}\bigg)^2=\sqrt{2\times9.8\frac{\mbox{m}}{\mbox{s}^2}\times50\mbox{m}}\times\frac{\sqrt{2\times9.8\frac{\mbox{m}}{\mbox{s}^2}\times50\mbox{m}}-3\frac{\mbox{m}}{\mbox{s}}}{2\frac{\mbox{m}}{\mbox{s}^2}}-\frac{1}{2}\times2\frac{\mbox{m}}{{s}^2}\times\bigg(\frac{\sqrt{2\times9.8\frac{\mbox{m}}{\mbox{s}^2}\times50\mbox{m}}-3\frac{\mbox{m}}{\mbox{s}}}{2\frac{\mbox{m}}{\mbox{s}^2}}\bigg)^2=242.75\mbox{m}$$ We can derive the full height keeping in mind that the height, at which the fall began, consists of deceleration phase $H_{dec}$ and free falling phase $H_{freely}$ $$H=H_{dec}+H_{freely}=242.75\mbox{m}+50\mbox{m}=292.75\mbox{m}$$