A ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the ground is

Take upward as positive. The height from the ground after time $t$ is $$h+vt-\frac{1}{2}gt^2=0.$$ So $$\frac{1}{2}gt^2-vt-h=0.$$ Using the quadratic formula, $$t=\frac{v\pm\sqrt{v^2+2gh}}{g}.$$ Since time must be positive, we take the positive root: $$t=\frac{v+\sqrt{v^2+2gh}}{g}.$$ This can be written as $$t=\frac{v}{g}\left(1+\sqrt{1+\frac{2gh}{v^2}}\right).$$ Comparing with the printed options, the intended matching choice is $\mathrm{B}$.