Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 150: 78

$40.1ft/s$

We can determine the required speed as follows: $\Sigma F_n=m\frac{v^2}{\rho}$ $\implies W=m\frac{v^2}{\rho}$ This can be rearranged as: $v=\sqrt{\frac{W\rho}{m}}$ $\implies \sqrt{\frac{mg \rho}{m}}$ $\implies v=\sqrt{\rho g}$ We plug in the known values to obtain: $v=\sqrt{(32.2)(50)}=40.1ft/s$