A ball rolls horizontally off the top of a stairway with a speed of $\mathbf{1}\mathbf{.}\mathbf{52}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. The steps are$\mathbf{20}\mathbf{.}\mathbf{3}\mathbf{}\mathit{c}\mathit{m}$high and 20.3 cmwide. Which step does the ball hit first?

As we know $\left(v_{0y}=0\right)$so, the coordinates of ball at time is given by $x=v_{0x}t\&y=-\frac{1}{2}g{t}^2$

Height of step is and the width is 20.3 cm.

We have been given height h and width w of the step. If we assume that the ball hits the n^th step, the ball has to travel nhvertically. nwwould be the length of the steps in the horizontal direction. So, the ball would hit between (n-1)wanddistance in terms of the length of the steps.

nw We take the origin of a coordinate system to be the point where ball leaves the top of the stairways. We choose upward as positive.

Formulae:

$s=v_{0}t+\frac{1}{2}g{t}^2$

Equate y = -nh and solve it for time then we can get time to reach the level of step n.

$t=\sqrt{\frac{2nh}{g}}$

The x coordinate is,

role="math" localid="1660899101305" $$\begin{gathered} x=v_{0x}\sqrt{\frac{2nh}{g}} \\ =(1.52m/s)\sqrt{2n\left(\frac{0.203m}{9.8m/s^2}\right)} \\ =(0.309m).\sqrt{n} \end{gathered}$$

We have the horizontal distance in terms of n, so we have to try different values if n for which it satisfy the required condition. The condition would be,

$n>\frac{x}{w}<\left(n-1\right)$

$n>\frac{(0.309\mathrm{m}).\sqrt{n}}{w}<\{n-1)$

x/w would give us number of steps. So the value of n can be found by putting the different values starting from 1.

As we choose the values of n=1, 2, 3… we get the answer at n=3 which satisfy above condition.

For$n=1,x=0.309mandx/w=1.52$, this is greater than n

For $n=2,x=0.437mandx/w=2.15$ this is also greater than n.

For $n=3,x=0.535mandx/w=2.64$Now, this is less than n and greater than n-1, so the ball hits the third step