Two projectiles A and B are thrown with initial velocities of $40 \mathrm{~m} / \mathrm{s}$ and $60 \mathrm{~m} / \mathrm{s}$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges respectively is $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

<p>The range of a projectile launched with an initial velocity $v$ at an angle $\theta$ with respect to the horizontal is given by:</p> <p>$R = \frac{v^2 \sin(2\theta)}{g}$,</p> <p>where $g$ is the acceleration due to gravity.</p> <p>Let&#39;s calculate the ranges of projectiles A and B:</p> <p>For projectile A, $v = 40 \, \text{m/s}$ and $\theta = 30^\circ$, so:</p> <p>$R_A = \frac{(40)^2 \sin(2 \times 30)}{10} = 4 \times 40 = 160 \, \text{m}$.</p> <p>For projectile B, $v = 60 \, \text{m/s}$ and $\theta = 60^\circ$, so:</p> <p>$R_B = \frac{(60)^2 \sin(2 \times 60)}{10} = 6 \times 60 = 360 \, \text{m}$.</p> <p>Therefore, the ratio of their ranges is $R_A : R_B = 160 : 360 = 4 : 9$.</p>