Answer: Option B
:
B
at t = 0, speed is 5 $\frac{m}{s}$, this means, u = 5 $\frac{m}{s}$. Also minimum speed is 3 $\frac{m}{s}$. For a projectile speed is minimum and equal to its horizontal component at the highest point of its trajectory, as the vertical component is zero at this point. Let $\theta$ be the angle of projection, then,
ucos $\theta$ = 3
$\Rightarrow$ 5 cos $\theta$ = 3
$\Rightarrow$ cos $\theta$ = $\frac{3}{5}$
$\theta$ = ${53}^{\circ }$
$\therefore$ Time of flight = $\frac{2usin\theta }{g}$ = $\frac{2\times 5\times sin{53}^{\circ }}{10}$
2$\times$5$\times$$\frac{4}{5}$$\times$ $\frac{1}{10}$
$\Rightarrow$ $\frac{4}{5}$s = 0.8s