How to calculate projectile off cliff when angle is below th

[Abcdefghijk]

Homework Statement

Cliff is 50 m tall
Stone thrown downwards with velocity of 12 m/s and 22 degrees below horizontal
What is the time it will take to hit the ground?

Homework Equations

V=d/t
D= Vi t + 1/2at^2

The Attempt at a Solution

Vx1= cos (22)(12)
= +11.12620625
Vy1= sin (22) (12)
= -4.495279121

I solved the triangle (hyp=12) but from here I am not sure what to do. Here is what I attempted from here:

Vertical
(Vi= 0 m/s, a (g)= -9.81 m/s^2, d= 50 m)
d= Vi t + 1/2at^2
t (re-arranged) = square root of d/(1/2a)
t= 10. 1936792 s

Horizontal
(V= 12 m/s, t= 10.1936792 s)
v= d/t
d= vt
d= (12)(10...)
d= 122.324159 m

The correct answer is 2.77 seconds, but I am unable to get this answer and comprehend how to achieve this answer. I would greatly appreciate a walkthrough/explanation of this question and if the angle was above the horizontal as well.

Thanks

 

[kuruman]

d= Vi t + 1/2at^2
t (re-arranged) = square root of d/(1/2a)
t= 10. 1936792 s

What happened to vi t? Isn't there a downward initial velocity component?

 

[Abcdefghijk]

For the vertical component, the initial velocity should be 0 m/s since the movement is horizontal. Therefore I think that you can eliminate Vi t as it will yield 0.

 

[kuruman]

Stone thrown downwards with velocity of 12 m/s and 22 degrees below horizontal

This means what is says. The stone is not shot horizontally. In fact you calculated its vertical component here

Vy1= sin (22) (12)
= -4.495279121

What do you think Vy1 is?