You are asking to use the general kinematic equation $S = ut + \frac{1}{2}at^2$ to solve for the horizontal distance, which was denoted as $R$ in the previous problem (and you are referring to as 'c'). Step 1: Apply the kinematic equation to horizontal motion. For horizontal motion, the displacement $S$ becomes the horizontal distance $R$, the initial velocity $u$ becomes the initial horizontal velocity $u_x$, and the acceleration $a$ becomes the horizontal acceleration $a_x$. $$ R = u_x t + \frac{1}{2}a_x t^2 $$ Step 2: Substitute the value for horizontal acceleration. In projectile motion, assuming no air resistance, the horizontal acceleration $a_x$ is $0 \, \text{m/s}^2$. $$ R = u_x t + \frac{1}{2}(0)t^2 $$ $$ R = u_x t $$ Step 3: Substitute the values from the previous problem. From the previous problem: Initial horizontal velocity $u_x = 10 \, \text{m/s}$ Time taken to reach the ground $t = \sqrt{10} \, \text{s}$ (calculated in part a) $$ R = (10 \, \text{m/s})(\sqrt{10} \, \text{s}) $$ $$ R = 10\sqrt{10} \, \text{m} $$ $$ R \approx 31.62 \, \text{m} $$ The horizontal distance (c) is $\boxed{\text{31.6 m}}$.