A cliff jumper runs off a 15 m high cliff. Rocks extend 3.0 m past the cliff bottom. We can ignore air resistance. What is the minimum horizontal velocity needed to clear the rocks?

The correct answer would be 1.7 m/s:

Start with what you know. In the y-direction, we know the jumper must fall 15 meters and starts with 0 velocity in the y direction. You can also assume that the acceleration of gravity is pulling down on the jumper at 9.8 m/s. Once you have these three you can plug it into kinematic a equation to find time

x=Vot+1/2at^2
15=(0)t+1/2(9.8)(t)^2
t=1.75

You get time=1.75 seconds. Since this is a kinematics problem, both the x and y direction have the same amount of time. You can then see that the x displacement is 3 to avoid the rocks, and acceleration is 0 in the x direction because no force is speeding it up. Therefore you can use the same equation to find initial velocity and final velocity, which are gonna be the same because we have 0 acceleration:

X=Vot+1/2at^2
3=Vi(1.75)+1/2(0)(1.75)^
Vi=1.7

1.7 is your answer

I have my work in the picture I really like to make charts to help keep everything organized if that helps you