Chapter 9 - Linear Momentum and Collisions - Problems and Conceptual Exercises - Page 293: 58

$840\frac{Kg}{s}$

We know that $thrust=(\frac{\Delta m}{\Delta t})(v)=mg$ This can be rearranged as: $\frac{\Delta m}{\Delta t}=\frac{mg}{v}$ We plug in the known values to obtain: $\frac{\Delta m}{\Delta t}=\frac{(5300)(9.81)}{62}$ $\frac{\Delta m}{\Delta t}=840\frac{Kg}{s}$