31. The height of a window of a bus traveling with uniform velocity is 0.5 m . Drops of paint are released simultaneously from the highest and lowest points of the window. The distance between the two drops of paint on the ground after hitting is 1 m . If the height of the lowest point of the window of the bus is 2 m from the ground, find the velocity with which the bus was traveling. (Ignore air resistance on paint drops.)

To solve this problem, we need to analyze the motion of the two drops of paint released from the window of the bus. 1. **Identify the heights of the drops:** - The height of the window is 0.5 m. - The lowest point of the window is 2 m from the ground. - Therefore, the highest point of the window is \(2 m + 0.5 m = 2.5 m\). 2. **Determine the time taken for each drop to hit the ground:** - The drop from the highest point (2.5 m) and the drop from the lowest point (2 m) will fall under the influence of gravity. - The time taken for an object to fall from a height \(h\) is given by the formula: \[ t = \sqrt{\frac{2h}{g}} \] where \(g \approx 9.81 \, \text{m/s}^2\). - For the drop from the highest point (2.5 m): \[ t_1 = \sqrt{\frac{2 \times 2.5}{9.81}} = \sqrt{\frac{5}{9.81}} \approx \sqrt{0.510} \approx 0.714 \, \text{s} \] - For the drop from the lowest point (2 m): \[ t_2 = \sqrt{\frac{2 \times 2}{9.81}} = \sqrt{\frac{4}{9.81}} \approx \sqrt{0.408} \approx 0.639 \, \text{s} \] 3. **Calculate the horizontal distance traveled by the bus:** - Let \(v\) be the velocity of the bus. - The horizontal distance traveled by the bus while the drops are falling is given by: \[ d = v \cdot t \] - The distance between the two drops on the ground is given as 1 m. The horizontal distance traveled by the drop from the highest point is: \[ d_1 = v \cdot t_1 \] and for the drop from the lowest point: \[ d_2 = v \cdot t_2 \] - The distance between the two drops on the ground is: \[ d_1 - d_2 = 1 \, \text{m} \] Substituting the expressions for \(d_1\) and \(d_2\): \[ v \cdot t_1 - v \cdot t_2 = 1 \] Factoring out \(v\): \[ v(t_1 - t_2) = 1 \] Therefore: \[ v = \frac{1}{t_1 - t_2} \] 4. **Calculate \(t_1 - t_2\):** \[ t_1 - t_2 \approx 0.714 \, \text{s} - 0.639 \, \text{s} \approx 0.075 \, \text{s} \] 5. **Calculate the velocity \(v\):** \[ v = \frac{1}{0.075} \approx 13.33 \, \text{m/s} \] Thus, the velocity with which the bus was traveling is approximately **13.33 m/s**.