A small disc of mass \(m\) slides down with initial velocity zero from the top \((A)\) of a smooth hill of height \(H\), having a horizontal portion \((BC)\) as shown in the figure. If the height of the horizontal portion of the hill is \(h\), then the maximum horizontal distance covered by the disc from the point \(D\) is
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For projectile motion after sliding from a smooth track, first use conservation of energy to find horizontal speed, then use vertical motion to find time of flight.
Step 1: Apply energy conservation from A to C. The disc starts from rest at height $H$ and slides (without friction) to the horizontal portion $BC$ at height $h$. By energy conservation: \[ mgH = mgh + \frac{1}{2}mv_C^2 \] \[ v_C^2 = 2g(H - h) \] where $v_C$ is the speed at the edge of the cliff (point D). Step 2: Set up the projectile motion after point D. After leaving D, the disc undergoes projectile motion. It has horizontal velocity $v_C$ and falls a vertical height $h$ before hitting the ground. Time to fall: \[ h = \frac{1}{2}gt_1^2 \implies t_1 = \sqrt{\frac{2h}{g}} \] Step 3: Write the horizontal distance from D. \[ x = v_C \cdot t_1 = \sqrt{2g(H-h)} \cdot \sqrt{\frac{2h}{g}} = \sqrt{4h(H-h)} = 2\sqrt{h(H-h)} \] Step 4: Maximize $x$ with respect to $h$. We maximize $f(h) = h(H-h) = Hh - h^2$. Taking derivative and setting to zero: \[ \frac{df}{dh} = H - 2h = 0 \implies h = \frac{H}{2} \] At $h = H/2$: \[ x_{\max} = 2\sqrt{\frac{H}{2} \cdot \frac{H}{2}} = 2 \cdot \frac{H}{2} = H \] Step 5: Confirm the maximum horizontal distance. At $h = H/2$: $v_C^2 = 2g(H - H/2) = gH$, and fall time $t_1 = \sqrt{2(H/2)/g} = \sqrt{H/g}$. So $x = \sqrt{gH} \cdot \sqrt{H/g} = H$. Step 6: State the final answer. The maximum horizontal distance covered by the disc from point D is $H$. \[ \boxed{H} \]
