Q:

How much is does it take for the stone to reach the bottom?

So a stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 65.0 m high. 
So, just making sure I have all my variables right 
Vinitial: 10 m/s 
V: 0 m/s 
g: -9.8 m/s^2 
y= 0 m 
y initial: 70.10 m 
I got 70.10 m by doing 10^2/2(9.8)= 5.10 and added that to the cliff's height because of the ball being thrown up in the air so that equals 70.10 m... is that correct? 
Also, I had to find how much later does it reach the bottom and tried using the quadratic equation for 1/2gt^2 + Vinitial(t) + y initial 
I got 2. something seconds but that wasn't the right answer. 
It was 4.8 s. What am I doing wrong???

A:

Hint:
$$y = y_0 + v_y t + \dfrac{1}{2}a_y t^2 =65+10t-\dfrac{1}{2}gt^2=0.$$
Use the quadratic formula.

