## College Physics (4th Edition)

Since the horizontal velocity is $3.0~m/s$, and the width of each step is $0.30~m$, the marble can move horizontally past one step each 0.1 seconds. Therefore, the marble can pass $n$ steps horizontally in a time of $(0.1~n)~seconds$ We can find the vertical displacement the marble falls in a time of $(0.1~n)$ seconds: $\Delta y = \frac{1}{2}at^2$ $\Delta y = \frac{1}{2}(9.8)(0.1~n)^2$ $\Delta y = (0.049~n^2)~meters$ We can find the total height of $n$ steps: $h = (0.18~n)~meters$ We need to find the lowest integer $n$ such that $\Delta y \gt h$: $(0.049~n^2) \gt (0.18~n)$ $(0.049~n) \gt (0.18)$ $n \gt \frac{0.18}{0.049}$ $n \gt 3.67$ $n = 4$ is the lowest integer such that $\Delta y \gt h$. Therefore, the marble strikes the fourth step first.