The vertical component of its initial velocity is
[tex]v_{0y}=\|\mathbf v_0\|\sin45^\circ=35\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Its height above the ground at time [tex]t[/tex] is given by
[tex]y=v_{0y}t-\dfrac g2t^2[/tex]
where [tex]g=9.8\,\frac{\mathrm m}{\mathrm s}[/tex] is the acceleration due to gravity. The ball reaches the ground when [tex]y=0[/tex], so for [tex]t>0[/tex] this occurs at
[tex]0\,\mathrm m=\left(35\,\dfrac{\mathrm m}{\mathrm s}\right)t-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2[/tex]
[tex]\implies t=7.1\,\mathrm s[/tex]
which makes B the most likely to be correct.
