# Question 09347

Feb 10, 2016

The ball hits the fifth step first.

#### Explanation:

The trick here is to realize that vertically, the number of steps traveled by the ball must be an integer. In other words, the distance traveled vertically must be a multiple of the height of a single step.

If we take $w$ to represent the height and width of a single step, we can say that

color(blue)(h) = n xx w" " " "color(red)(("*"))

Here

$\textcolor{b l u e}{h}$ - the height traveled by the ball
$n$ - the number of steps

Horizontally, on the other hand, the ball does not have to travel for a multiple of the width of a single step, since it can actually fall between two steps.

However, its position will be determined by $n$, the number of steps it falls vertically.

More specifically, its horizontal displacement will vary between $\left[\left(n - 1\right) \cdot w\right]$ and $\left[n \cdot w\right]$, as you can see in the above image.

Now, since the ball is thrown horizontally from the top of the stairs, its initial velocity will only have one component, the horizontal one, ${v}_{0 x}$.

The vertical component of the initial velocity, ${v}_{0 y}$, will be equal to zero.

Horizontally, the motion of the ball is unaffected by gravity, which is why ${v}_{0 x}$ is constant throughout its flight.

If we take $\textcolor{g r e e n}{d}$ to be the horizontal distance covered by the ball, we can say that

color(green)(d) = v_(0x) * t" " " "color(red)(("* *"))

Here

$t$ - the time of flight

Vertically, the motion of the ball is influenced by gravity. The ball will be accelerated by the gravitational acceleration, $g$. This means that you can write, for the vertical portion of the ball's motion

$h = {\overbrace{{v}_{0 y} \cdot t}}^{\textcolor{p u r p \le}{= 0}} + \frac{1}{2} \cdot g \cdot {t}^{2}$

$\textcolor{b l u e}{h} = \frac{1}{2} \cdot g \cdot {t}^{2}$

Remember, the time of flight is the same for both directions of movement!

Use equation $\textcolor{red}{\left(\text{*}\right)}$ to write

$n \cdot w = \frac{1}{2} \cdot g \cdot {t}^{2}$

Solve this for $t$ as a function of $n$

$2 \cdot n \cdot w = g \cdot {t}^{2} \implies t = \sqrt{\frac{2 \cdot n \cdot w}{g}}$

Now, take this expression for $t$ and plug it into the equation $\textcolor{red}{\left(\text{* *}\right)}$. This will help you find $\textcolor{g r e e n}{d}$ as a function of $n$

$\textcolor{g r e e n}{d} = {v}_{0 x} \cdot \sqrt{\frac{2 \cdot n \cdot w}{g}}$

$\textcolor{g r e e n}{d} = {v}_{0 x} \cdot \sqrt{\frac{2 \cdot w}{g}} \cdot \sqrt{n}$

Plug in your values to get

d = "2.3 m" color(red)(cancel(color(black)("s"^(-1)))) * sqrt( (2 * 25 * 10^(-2)color(red)(cancel(color(black)("m"))))/(9.81color(red)(cancel(color(black)("m")))"s"^(-2))) * sqrt(n)#

$d = \left(0.519 \cdot \sqrt{n}\right) \text{ m}$

Now, to find the value of $n$, you must use the fact that you have

$\left(n - 1\right) \cdot w < d < n \cdot w$

This is of course equivalent to saying that

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

For $n = 1$, you have

$1 - 1 < \frac{0.519 \cdot \sqrt{1}}{0.25} < 1$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{0 < 2.08 < 1}}} \to$ Not a valid solution

Do this until you come up with a suitable value of $n$. I ended up with $n = 5$, since

$4 < \frac{0.519 \cdot \sqrt{5}}{0.25} < 5$

$4 < 4.642 < 5 \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$