Question

A staircase contains three steps each $10cm$ high and $20cm$ wide. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as it hits directly the lowest plane?

Answer 1

The vertical and horizontal distance covered by the body will be $2h$ and $2b$ respectively. You can start by using the second equation of motion, i.e. $h = ut + \dfrac{1}{2}a{t^2}$ for the vertical motion of the ball and calculate the value of $t$ . Then put this value of $t$ in the equation for the speed of the horizontal motion of the body, i.e. $s = \dfrac{d}{t}$ .

Complete answer:
In the problem given to us
Height of each step of the staircase, $h = 10cm$
Width of each step of the staircase, $b = 20cm$

Let the initial speed of the ball and time taken by the ball to hit the lowest plane be $u$ and $t$ respectively.
The total horizontal distance that the ball has to cover is equal to the width of two steps of the staircase combined $= 2b = 2 \times 20 = 40cm = 0.4m$
We know that the equation for speed is
$s = \dfrac{d}{t}$
$s =$ Speed of the ball
$d =$ The horizontal distance that the ball covers
$t =$ The time taken by the ball to cover the distance.
This equation can be used to calculate the horizontal speed of the equation as follows
$u = \dfrac{{2b}}{t}$
$u = \dfrac{{0.4}}{t}$
$t = \dfrac{{0.4}}{u}$ (Equation 1)
Now it becomes clear that to calculate the horizontal speed of the ball we have to calculate the time taken by the ball to hit the lowest stair.
Total vertical distance by the ball covers
We know that the second equation of motion is
$h = ut + \dfrac{1}{2}a{t^2}$
Here, $h =$ Height
$u =$ The initial speed of the ball
$t =$ Time
$a = 9.8m/{s^2} =$ The acceleration due to gravity
So, for the vertical motion of the ball
$0.2 = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
${t^2} = \dfrac{{0.4}}{{9.8}}$
${t^2} = 0.04$
$t = 0.2\sec$
Now, substituting the value of $t$ in equation 1 we get
$0.2 = \dfrac{{0.4}}{u}$
$u = \dfrac{{0.4}}{{0.2}}$
$u = 2m/s$
Hence, the ball must have a minimum horizontal velocity of $2m/s$ to hit the lowest step of the staircase without hitting the steps in the middle.

Note:
In the problem given to us we had to calculate the minimum horizontal velocity of the ball to hit the lowest step of the staircase without hitting the steps in the middle. If you had to calculate the maximum horizontal velocity the steps of calculation would remain the same but you would take horizontal as $3b$ instead of $2b$.