Flight time of Projectile motion

[wassup]

Ok, I thought I knew how to do this, but for some reason, according to my homework website, my answer is wrong. So I'm turning to this board before I beat my head against the wall. Please help me.

Homework Statement

A ball is thrown with an initial speed of 31 m/s at an angle of 45�. The ball is thrown from a height of 12 m and lands on the ground.
Find the time of flight.

The Attempt at a Solution

Initial vertical upward velocity = 31 sin 45 = 26.378m/s.
The displacement [Yf – Yo] = –12 m
Yf -Yo = Vot - 0.5*g*t^2
–12 = 26.378 t – 4.905 *t^2
4.905 *t^2 – 26.378 t – 12 = 0
--------------------------------------...
Using b^2 – 4ac = 26.378 ^2– 4*4.905*(– 12)
√ (b^2 – 4ac) = 30.516
– b ±√ (b^2 – 4ac) = 26.378 ± 30.516
= 56.894 leaving the negative time value
Dividing by 2a = 9.81
t = 5.7996 s

Is my work right?

 

[LowlyPion]

Welcome to PF.

I'd check the vertical velocity component again.

31 * sin45° = ... ?

 

[wassup]

Ugh, no wonder none of my answers have been right, I've been in radian mode on my calculator. I feel like an idiot. Thank you for pointing out the obvious to me.

 

[LowlyPion]

Here's a useful link for your quadratic:
http://www.math.com/students/calculators/source/quadratic.htm