91Ó°ÊÓ

The ball falls with projectile motion due to gravity. Let's first calculate the time it takes to fall each vertical distance. The first step is to use the formula for vertical motion \[ y = \frac{1}{2} g t^2 \]where \(y\) is the vertical distance, and \(g = 9.8 \text{ m/s}^2\) is acceleration due to gravity. For each vertical step \(0.2 \text{ m}\), solve for \(t\).* Step 1 vertical calculation: Substitute \(y = 0.2\) into the equation: \[ 0.2 = \frac{1}{2} \times 9.8 \times t^2 \] \[ t^2 = \frac{0.2 \times 2}{9.8} \] \[ t^2 = \frac{0.4}{9.8} \approx 0.0408 \] \[ t \approx 0.202 \text{ s} \]

We calculate how far the ball travels horizontally in this time using the formula for horizontal distance because velocity in the horizontal direction is constant:\[ x = v_x t \]where \(v_x = 1.8 \text{ m/s}\).For step 1:\[ x_{1} = 1.8 \times 0.202 \approx 0.3636 \text{ m} \]So, for step 1, the ball travels \(0.3636\) meters horizontally.

Each step has a width of \(0.2 \text{ m}\). We need to check if the horizontal distance covered for each corresponding time calculated in Step 1 exceeds the width of the step.

For the first step, the distance \(0.3636 \text{ m}\) exceeds the horizontal width \(0.2 \text{ m}\), indicating that the ball will skip over it since the ball will land further than the first step width.Perform similar calculations for the second step:1. Time to fall the height of two steps: \[ y = 0.4 \text{ m} \] \[ 0.4 = \frac{1}{2} \times 9.8 \times t^2 \] \[ t^2 = \frac{0.8}{9.8} \approx 0.0816 \] \[ t \approx 0.286 \text{ s} \]2. Horizontal distance: \[ x_{2} = 1.8 \times 0.286 \approx 0.5148 \text{ m} \] Since \(0.5148 \text{ m} > 0.4 \text{ m}\), it will skip the second step as well.Now for the third step:3. Time to fall the height of three steps: \[ y = 0.6 \text{ m} \] \[ 0.6 = \frac{1}{2} \times 9.8 \times t^2 \] \[ t^2 = \frac{1.2}{9.8} \approx 0.1224 \] \[ t \approx 0.350 \text{ s} \]4. Horizontal distance: \[ x_{3} = 1.8 \times 0.350 \approx 0.630 \text{ m} \] The distance \(0.630 \text{ m} > 0.6 \text{ m}\) exceeds the next step's foot, so it skips the third step.For the fourth step, let’s perform similar calculations, but since the previous results underestimate the specific step the ball will fall on after further horizontal consideration, the next detailed calculations involve analyzing exact positioning given previous approximations.