To solve this problem, we will use the kinematic equations for projectile motion.$\\ \\$Given:$\\$• Initial speed of the bullet $u = 280 \,$m/s$. \\$• Angle of projection $\theta = 30^\circ. \\$• Gravitational acceleration $g = 9.8 \,$m/s$^2. \\$• $\sin 30^\circ = 0.5. \\ \\$The vertical component of the initial velocity $u_y$ is given by:$\\ u_y = u \sin \theta \\ u_y = 280 \times 0.5 \\ u_y = 140 \,$m/s$. \\ \\$The maximum height $H$ attained by the bullet is given by the equation:$\\ v_y^2 = u_y^2 - 2gH \\$At maximum height, the final vertical velocity $v_y = 0. \\ 0 = u_y^2 - 2gH \\ 2gH = u_y^2 \\ H = \frac{u_y^2}{2g} \\ \\$Substitute the known values:$\\ H = \frac{140^2}{2 \times 9.8} \\ H = \frac{19600}{19.6} \\ H = 1000 \,$m$. \\ \\$Therefore, the maximum height attained by the bullet is $1000 \,$m$. \\$This corresponds to Option 3.