Physics Help Finding Distance with velocity

Question

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 5 m at your feet, then a horizontal shelf of 9 m, then another drop of 5 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.55 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s 2. Consider air friction to be negligible. Answer in units of m/s.

I got 8.91 for the horizontal velocity (part 1/2).

I cannot figure out part 2 to find the distance of the 2nd cliff question.

Answer 1

Part 1: To find the initial horizontal velocity, you need to calculate the time it would take to reach the edge of the 9 meter shelf. It turns out that that time is the same time to fall the 5 meter distance to the shelf. This can be found using the equation:

H = Vot + ½ g t^2 where H is 5 meters and Vo = 0 (no initial downward velocity).

Substitute the known variables and solve for t:

t = √ (2 x H / 9.8) = √ (2 x 5 / 9.8) = 1.01 seconds.

Now use t in the equation Vh = 9m / 1.01 s = 8.9 m/s where Vh is the initial horizontal velocity.

Note that the mass of the rock doesn’t come into play in this part.

I don’t see part 2.