Ball Rolling Horizontally Off a Staircase

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Kinematics Intermediate projectile

Source: Principles of Physics

Problem Sets:

Kinematics review1 Kinematics set 1

Problem

A ball rolls horizontally off the top of a stairway with an initial speed $v_0$. The steps each have height $h$ and width $w$.

Which step does the ball hit first?

The ball first hits step $n$, where $n$ is the smallest integer satisfying the condition:

$$n > \frac{2v_0^2 h}{gw^2}$$

Let the origin be the edge of the top step, with the x-axis horizontal and the y-axis vertically downward. The ball's position is given by the equations of projectile motion:

$$x(t) = v_0 t$$ $$y(t) = \frac{1}{2}gt^2$$

By eliminating the time variable $t = x/v_0$, we find the ball's trajectory equation:

$$y(x) = \frac{g}{2v_0^2}x^2$$

The top-front corner of the $n$-th step is located at the coordinates $(nw, nh)$. The ball will land on or beyond the $n$-th step if its trajectory passes below this corner. This means that at the horizontal position $x = nw$, the ball's vertical position $y$ must be greater than $nh$.

$$y(x=nw) > nh$$

Substituting into the trajectory equation gives the condition:

$$\frac{g}{2v_0^2}(nw)^2 > nh$$

Assuming $n>0$, we can simplify and solve for $n$:

$$\frac{gnw^2}{2v_0^2} > h$$ $$n > \frac{2v_0^2 h}{gw^2}$$

The step number must be an integer. The first step the ball can possibly hit is the smallest integer $n$ that satisfies this inequality.