A package falls out of an airplane which is flying horizontally at a constant velocity $\left(u\right)$ and fixed height $\left(h\right)$. If air resistance is ignored, then which of the following is the expression of magnitude of velocity $\left(v\right)$ of package after time $t?$

The correct option is A $v=\sqrt{u^2+g^2{t}^2}$
Let $x-$axis denotes horizontal direction and $y-$axis denotes vertical downward direction.

At the projection,
$u_x=u;u_y=0$

$\therefore \overrightarrow{u}=u\hat{i}$

And, after time $t$, velocity of package will be,

$\Rightarrow \overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

As there is no acceleration in horizontal direction, so

$v_x=u$

And acceleration due to gravity $\left(g\right)$ in the vertical downward direction, velocity in downward direction will be,

$v_y=u_y+gt$

$v_y=gt(\because u_y=0)$

So, the final velocity of package after time $t$ will be,

$\overrightarrow{v}=\left(u\right)\hat{i}+\left(gt\right)\hat{j}$

$\therefore |\overrightarrow{v}|=\sqrt{u^2+g^2{t}^2}$

Hence, option $\left(\text{B}\right)$ is the correct answer.