Question 1.23: An airplane is 20,000 feet above the ground when a 100 kg ob...

\begin{array}{ll}m g \Delta z+\frac{m \Delta \nu^{2}}{2}=0\\\frac{\Delta \nu^{2}}{2}=-g \Delta z\\0.5\left(\nu_{2}^{2}-\nu_{1}^{2}\right)=-g\left(z_{2}-z_{1}\right), \nu_{1}=0, z_{2}=0\\\nu_{2}=\sqrt{2 g z_{1}}\\\nu_{2}=\sqrt{(2)\left(\frac{32.2 \,f t}{s^{2}}\right)(20000 \,f t)}=1135 \,ft / \mathrm{s}\\K . E .=\frac{m \nu^{2}}{2}=\frac{(100 \mathrm{~kg})(1135 \,ft / \mathrm{s})^{2}}{2}\left\lgroup\frac{0.3048 ~m}{1 \,ft}\right\rgroup^{2}\left\lgroup\frac{1 ~N}{1 ~kgm / s^{2}}\right\rgroup\left\lgroup\frac{1 ~J}{Nm}\right\rgroup=5.984 \times 10^{6} ~J\\\bf K.E. =5 . 9 8 4 \times 10^{6} J\end{array}