I am given that a projectile is launched from ground level and lands at a distance L below ground level.
It is launched at an angle $\theta$ with a velocity V.
I have found using F=ma that the position vector at time $t$ is $\mathbf r=(Vcos\alpha)t \ \hat j + ((Vsin\alpha)t-\frac{gt^2}{2})\hat k$.
In order to calculate the time of flight am I right in setting $Vsin\alpha t-\frac{gt^2}{2}=L$ and then solving for t?
As michael has already said, the displacement is a vector quantity and can be negative.
Since you have considered acceleration vector to be negative and velocity vector to be positive, and final position is below ground or below $y=0$, the displacement in $y$ direction will be negative.
So the equation will have $-L$ instead of $L$