A ball rolls off the top of a stairway with a horizontal velocity of magnitude \(5.0 \mathrm{ft} / \mathrm{s}\). The steps are \(8.0\) in. high and \(8.0\) in. wide. Which step will the ball hit first?

To calculate this, we consider that the vertical displacement is equal to the height of one step which is \(8.0\) in. We may need to convert this height into feet for consistency with the given horizontal velocity. One foot equals 12 inches, so a height of \(8.0\) in. can be written as \( \frac{8.0}{12} \) feet or \( \frac{2}{3} \) feet. The ball falls under the influence of gravity, therefore consider acceleration \(g = 32.2 \, \text{ft/s}^2\). Since initial vertical velocity is 0, use the equation of motion \(y = v_i t + \frac{1}{2} a t^2\) where \(y\) is the vertical displacement, \(v_i\) is the initial vertical velocity, \(t\) is time, and \(a\) is acceleration due to gravity. From this, \(t = \sqrt{\frac{2y}{g}}\).

In projectile motion, the horizontal velocity remains constant. Therefore, the horizontal distance can be calculated by the product of time and the given horizontal velocity, \(v_h = 5.0 \, \text{ft/s}\). Use the equation of motion in a horizontal direction, \(x = v_h \cdot t\).

The problem becomes simple if the ball moves horizontally the same distance that it does vertically which means the ball will hit step 1. If the horizontal displacement is larger than the width of a step, then the ball will hit a following step. Find out the step that the ball will hit by dividing the horizontal displacement by the width of a step which is also \(8.0\) in or \( \frac{2}{3} \) ft. The result will give which step the ball will hit first.