To determine which step the ball hits first, we need to analyze its motion after it rolls off the top of the stairway. The ball has an initial horizontal velocity and falls freely under gravity. We will apply the concepts of projectile motion to solve this problem.
- The horizontal velocity (\(v_x\)) of the ball is given as \(1.8 \, \text{m/s}\). The horizontal distance covered by the ball in time \(t\) is given by: \(x = v_x \cdot t\)
- The vertical motion is described by the acceleration due to gravity (\(g = 9.8 \, \text{m/s}^2\)). The vertical distance covered in time \(t\) is given by: \(y = \frac{1}{2}gt^2\)
- Each step is \(0.20 \, \text{m}\) high and \(0.20 \, \text{m}\) wide. This setup defines a coordinate space where: - At the first step, \(x = 0.20 \, \text{m}\) and \(y = 0.20 \, \text{m}\) - At the second step, \(x = 0.40 \, \text{m}\) and \(y = 0.40 \, \text{m}\) - At the third step, \(x = 0.60 \, \text{m}\) and \(y = 0.60 \, \text{m}\) - At the fourth step, \(x = 0.80 \, \text{m}\) and \(y = 0.80 \, \text{m}\)
- To find the time the ball takes to fall to each step, solve the equation \(y = \frac{1}{2}gt^2\) for \(t\):
\(t = \sqrt{\frac{2y}{g}}\) - For each step, calculate the corresponding \(t\) and the horizontal distance \(x\) the ball would travel in that time:
- For the first step:
\(t_1 = \sqrt{\frac{2 \times 0.20}{9.8}} \approx 0.20 \, \text{s}\)
\(x_1 = 1.8 \times 0.20 = 0.36 \, \text{m}\) (The ball needs \(x_1 = 0.20 \, \text{m}\)) - For the second step:
\(t_2 = \sqrt{\frac{2 \times 0.40}{9.8}} \approx 0.29 \, \text{s}\)
\(x_2 = 1.8 \times 0.29 = 0.52 \, \text{m}\) (The ball needs \(x_2 = 0.40 \, \text{m}\)) - For the third step:
\(t_3 = \sqrt{\frac{2 \times 0.60}{9.8}} \approx 0.35 \, \text{s}\)
\(x_3 = 1.8 \times 0.35 = 0.63 \, \text{m}\) (The ball needs \(x_3 = 0.60 \, \text{m}\)) - For the fourth step:
\(t_4 = \sqrt{\frac{2 \times 0.80}{9.8}} \approx 0.40 \, \text{s}\)
\(x_4 = 1.8 \times 0.40 = 0.72 \, \text{m}\) (The ball needs \(x_4 = 0.80 \, \text{m}\))
- The ball will hit the first step where the horizontal distance \(x\) calculated matches or exceeds the width needed.
The ball first meets this criterion at the fourth step. Therefore, the ball will hit the fourth step first.
Hence, the correct answer is Fourth.