# Chapter 5 - Exercises and Problems: 28

$8.46 \ m/s$

#### Work Step by Step

We know the y-forces must cancel, for the ball remains in the horizontal plane. Thus, we find: $mg = F_t sin12$ We also know that the x-component of tension is the centripetal force: $F_tcos12 =\frac{mv^2}{r}$ Thus, we use substitution to find: $mg/tan12=\frac{mv^2}{r}$ $v=\sqrt{grtan12}=\sqrt{\frac{9.81\times1.55}{tan12}}=8.46 \ m/s$

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