# Question #09347

##### 1 Answer

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

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

Here

**height** traveled by the ball

**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

More specifically, its horizontal displacement will vary between

Now, since the ball is thrown **horizontally** from the top of the stairs, its initial velocity will only have **one** component, the *horizontal one*,

The vertical component of the initial velocity, **equal to zero**.

**Horizontally**, the motion of the ball is unaffected by gravity, which is why **constant** throughout its flight.

If we take

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

Here

*time of flight*

**Vertically**, the motion of the ball is influenced by gravity. The ball will be accelerated by the gravitational acceleration,

#h = overbrace(v_(0y) * t)^(color(purple)(=0)) + 1/2 * g * t^2#

#color(blue)(h) = 1/2 * g * t^2#

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

Use equation

#n * w = 1/2 * g * t^2#

Solve this for

#2 * n * w = g * t^2 implies t = sqrt((2 * n * w)/g)#

Now, take this expression for

#color(green)(d) = v_(0x) * sqrt((2 * n * w)/g)#

#color(green)(d) = v_(0x) * sqrt((2 * w)/g) * 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 = (0.519 * sqrt(n))" m"#

Now, to find the value of

#(n-1) * w < d < n * w#

This is of course equivalent to saying that

#(n-1) < d/w < n#

For

#color(red)(cancel(color(black)(0 < 2.08 < 1))) -># Nota valid solution

Do this until you come up with a suitable value of

#4 < (0.519 * sqrt(5))/0.25 < 5#

#4 < 4.642 < 5color(white)(a)color(green)(sqrt())#