Person A is standing 9m above the ground on a treehouse. Person B throws a ball upwards at 14m/s. How long does Person A have to wait to catch the ball if they missed it coming up?

The equation for vertical motion is s(t)=-4.9t^2+v_0t+s_0 where s(t) is the height in meters at time t in seconds, v(0) is the initial velocity, and s(0) is the initial height.
The height we are interested in is 9 m. The initial velocity is 14 m/s. We will take the initial height to be 0 m. (Though it would make more sense to have thrown the ball from some height like 2 m, since this is roughly the average height of a person, many physics problems simplify this to make the arithmetic easier.)
Then we have 9=-4.9t^2+14t ==> -4.9t^2+14t-9=0
Using the quadratic equation we get two answers: t~~.9768, t~~1.8803
The first is the time from the throw (when the ball is coming up); the second is the time as the ball falls back down.
Thus the person in the tree has about 1.88 seconds to catch the ball.