A freely falling stone crashes through a horizontal glass plate at time \(t\) and losses half of its velocity. After time \((t / 2)\) it falls on the ground. The glass plate is \(60 \mathrm{~m}\) high from the ground. Find the total distance travelled by the stone. \(\left[g=10 \mathrm{~ms}^{-2}\right]\) (A) \(120 \mathrm{~m}\) (B) \(80 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(140 \mathrm{~m}\)

We are given that the stone falls on the ground after time \(t\) when it crashes through the glass plate. The glass plate is \(60\mathrm{~m}\) high from the ground, and as it takes \((t / 2)\) to fall to the ground, it takes \((t - t / 2) = (t / 2)\) to fall through the glass plate. So, we have: \[t_\text{glass} = \frac{t}{2}\]

Let's represent the velocity of the stone when it hits the glass plate as \(v_\text{glass}\). When the stone passes through the glass plate, it loses half of its velocity. We can write this relationship as \(v_\text{after} = 0.5v_\text{glass}\). We know from the given exercise that the time taken to fall to the ground from the glass plate is equal to the time taken to fall through the glass plate, that is, \(t_\text{glass} = t / 2\). We can use the equation of motion: \[v_\text{after} = v_\text{glass} - g \cdot t_\text{glass}\] Substitute \(v_\text{after} = 0.5v_\text{glass}\) and \(t_\text{glass} = t / 2\): \[0.5v_\text{glass} = v_\text{glass} - 10 \cdot \frac{t}{2}\]

Now, let's use the second equation of motion to find the initial velocity of the stone, \(v_\text{glass}\): \[s_\text{glass} = v_\text{glass} \cdot t_\text{glass} - \frac{1}{2}gt_\text{glass}^2\] We are given the plate's height \(s_\text{glass} = 60\mathrm{~m}\): \[60 = v_\text{glass} \cdot \frac{t}{2} - \frac{1}{2}\cdot 10 \cdot \left(\frac{t}{2}\right)^2\]

Solve the two equations for \(v_\text{glass}\) and \(t\): \[0.5v_\text{glass} = v_\text{glass} - 10 \cdot \frac{t}{2}\hspace{1cm}(1)\] \[60 = v_\text{glass} \cdot \frac{t}{2} - \frac{1}{2}\cdot 10 \cdot \left(\frac{t}{2}\right)^2\hspace{1cm}(2)\] Solving (1) for \(t\), we get: \[t = 5\mathrm{s}\] Plugging the \(t = 5\mathrm{s}\) in (2), we get: \[v_\text{glass} = 40\mathrm{~ms}^{-1}\]

Now we will find the distance traveled during each phase separately and then add those distances to find the total distance traveled by the stone. Phase 1: Distance traveled until the stone hits the glass plate is \(s_\text{glass} = 60\mathrm{~m}\). Phase 2: Distance traveled after the stone passes through the glass plate is \(s_\text{after} = v_\text{after} \cdot t_\text{glass} = 0.5v_\text{glass} \cdot t_\text{glass} = 0.5(40\mathrm{~ms}^{-1})(2.5\mathrm{s})= 50\mathrm{~m}\). So, the total distance traveled by the stone is: \[s_\text{total} = s_\text{glass} + s_\text{after} = 60\mathrm{~m} + 50\mathrm{~m} = 110\mathrm{~m}\] Thus, none of the given options match the obtained result, but the closest option to the calculated total distance traveled is (C) \(100\mathrm{~m}\).