The ball hits the fifth step first.
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)(("*"))#
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,
Horizontally, the motion of the ball is unaffected by gravity, which is why
If we take
#color(green)(d) = v_(0x) * t" " " "color(red)(("* *"))#
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!
#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#
#color(red)(cancel(color(black)(0 < 2.08 < 1))) ->#Not a 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())#