The question has a link to it, please read it. Thanks.

http://www.phy.duke.edu/~rgb/Class/review53/node10.html

First, use the formula d = 1/2 * a * t^2

Solving for t gives t = ( 2 * d / a) ^ 1/2 (i.e., the square root of 2d over a)

So, t = 29.97 sec ... that's how long it takes to hit the ground.

Now, the ball travels horizontally at 120 m/s for 29.97 seconds, so it
hits at 120*29.97 or 3596 meters from the base of the cliff.

The last question is trickier ... the velocity has a horizontal and a
vertical component. The horizontal component is, of course, the 120
m/s.

The vertical component is given by v = a * t, or 1.67 * 29.97 which
equals 50.05 m/s.

The magnitude of the resultant vector is the square root of 120
squared plus 50.05 squared ... or 130.0 m/s.