Answer:
To solve this problem, we can use the kinematic equation for horizontal motion, which relates the initial velocity ($v_{0}$), final velocity ($v_{f}$), acceleration ($a$), and displacement ($d$) of an object:
$d = v_{0} t + \frac{1}{2}at^{2}$
In this case, we want to find the minimum initial velocity ($v_{0}$) that the divers must have to clear the rock. To do this, we can assume that the divers just graze the rock at the start of their trajectory, so the displacement in the horizontal direction is equal to the distance from the cliff to the rock ($d = 9.34 m$). We also know that the acceleration in the horizontal direction is zero, so the only force acting on the divers is gravity in the vertical direction, which gives an acceleration of $a = 9.8 m/s^{2}$.
At the instant the divers leave the cliff, they have zero horizontal velocity, so $v_{0} = 0$. We can use the equation above to solve for the time it takes for the divers to fall from the cliff to the level of the rock:
$d = \frac{1}{2}at^{2} \Rightarrow t = \sqrt{\frac{2d}{a}}$
Plugging in the numbers, we get:
$t = \sqrt{\frac{2(9.34 m)}{9.8 m/s^{2}}} \approx 1.44 s$
Since the cliff divers want to clear the rock, they need to travel a horizontal distance of at least $9.34 m$ during this time. We can use the equation for horizontal motion again to solve for the minimum initial velocity:
$d = v_{0}t \Rightarrow v_{0} = \frac{d}{t} = \frac{9.34 m}{1.44 s} \approx 6.49 m/s$
Therefore, the minimum horizontal velocity that the cliff divers must have to clear the rock is approximately $6.49 m/s$.