KCET 2025 Physics Paper

To solve this problem, we need to determine the time, denoted as $t_0$, at which the separation between two falling stones becomes H . The second stone starts falling $\Delta t$ seconds after the first stone.Distance Fallen by the First Stone ( $S_1$ ): The first stone falls for $t_0$ seconds, so the distance it covers is given by:$S_1 = \frac{1}{2} g t_0^2$Distance Fallen by the Second Stone $(S_2)$ :The second stone starts falling $\Delta t$ seconds after the first, so it falls for $(t_0 - \Delta t)$ seconds. Thus, the distance it covers is:$S_2 = \frac{1}{2} g (t_0 - \Delta t)^2$ Distance of Separation (H):The separation distance between the two stones is:$H = S_1 - S_2 = \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0 - \Delta t)^2$Simplifying the equation:$H = \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0^2 - 2 t_0 \Delta t + (\Delta t)^2)$$= \frac{1}{2} g \cdot (2 t_0 \Delta t - (\Delta t)^2)$$= g \cdot \left(t_0 \Delta t - \frac{1}{2} \Delta t^2\right)$ Expression for $t_0$ :Solving the above equation for $t_0$, we have:$H = g \left(t_0 \Delta t - \frac{1}{2} \Delta t^2\right)$$t_0 \Delta t = \frac{H}{g} + \frac{1}{2} \Delta t^2$$t_0 = \frac{H}{g \Delta t} + \frac{\Delta t}{2}$ Therefore, the time $t_0$ at which the separation between the two stones becomes H is given by:$t_0 = \frac{H}{g \Delta t} + \frac{\Delta t}{2}$