## Physics Friday 119

A skydiver of mass 125 kg (including gear), jumps from a plane, travelling with an air speed of 40 m/s at an altitude of 4000 meters. By the time they have descended to 3400 meters, his velocity is sufficiently close to the terminal velocity of 55 meters per second that the further aymptotic approach may be neglected, and velocity approximated by that 55 m/s. After another 45 seconds, he pulls the cord for the parachute. What is the total heat dissipated into the environment (and the skydiver) by the air resistance in the time period between leaving the plane and pulling the cord?

The simplest method is to consider energy conservation; the heat dissipated is equal to the mechanical energy lost by the skydiver. When he leaves the plane, his energy is sum of the (horizontal) kinetic energy and the gravitational potential energy $\begin{eqnarray}E_i&=&T_i+V_i\\&=&\frac12mv^2+mgh\\&=&\frac12(125\,\text{kg})\left(40\,\frac{\text{m}}{\text{s}}\right)^2+(125\,\text{kg})\left(9.8\,\frac{\text{m}}{\text{s}^2}\right)(4000\,\text{m})\\&=&5.0\times10^6\,\text{J}\end{eqnarray}$.
Now, if he falls for 45 seconds at 55 m/s, he pulls the cord at an altitude of (3400 m)-(55 m/s)(45 s)=925 m. Thus, his mechanical energy then is $\begin{eqnarray}E_f&=&T_f+V_f\\&=&\frac12mv^2+mgh\\&=&\frac12(125\,\text{kg})\left(55\,\frac{\text{m}}{\text{s}}\right)^2+(125\,\text{kg})\left(9.8\,\frac{\text{m}}{\text{s}^2}\right)(925\,\text{m})\\&=&1.3\times10^6\,\text{J}\end{eqnarray}$,
and thus the heat is
5.0×106-1.3×106=3.7×106 Joules.