#### Answer

The speed at the bottom of the dip is 12.1 m/s

#### Work Step by Step

Let $F_N$ be the normal force exerted on the passengers. Note that $F_N$ is equal to the apparent weight of the passengers. We can let $F_N = 1.5~mg$ to find the speed at the bottom of the dip.
$\sum F = \frac{mv^2}{r}$
$F_N-mg = \frac{mv^2}{r}$
$1.5~mg-mg = \frac{mv^2}{r}$
$0.5~g = \frac{v^2}{r}$
$v = \sqrt{0.5~g~r}$
$v = \sqrt{(0.5)(9.80~m/s^2)(30~m)}$
$v = 12.1~m/s$
The speed at the bottom of the dip is 12.1 m/s.