#### Answer

The player was running at a speed of 5.9 m/s

#### Work Step by Step

The amount of thermal energy is equal to the player's initial kinetic energy;
$KE = E_{th}$
$\frac{1}{2}mv^2 = 950~J$
$v^2 = \frac{(2)(950~J)}{m}$
$v = \sqrt{\frac{(2)(950~J)}{m}}$
$v = \sqrt{\frac{(2)(950~J)}{55~kg}}$
$v = 5.9~m/s$
The player was running at a speed of 5.9 m/s.