Question

A ball rolls horizontally off the top of a stairway with a speed of $1.5{\text{ms}}^{-1}$. The steps are $0.2\text{m}$ high and $0.2\text{m}$ wide. If the balls hits the edge of $n^{th}$ step first, then $n$ is equal to

Answer 1

Total height through which the ball will fall down vertically before it strikes the $n^{th}$ step $h=0.2\times n=0.2n$
Using second equation of motion, we get
$h=\frac{1}{2}g{t}^2$
$0.2n=5t^2\text{(i)}$
Horizontal distance travelled by the ball before striking the $n^{th}$ step
$R=0.2\times n=0.2n$
Now, $0.2n=u_x\times t$
$\Rightarrow 0.2n=1.5\times t\text{(ii)}$
Equating (i) and (ii), we get
$5t^2=1.5t\Rightarrow t=0.3$
Substituting in (i), $n=5\times {0.3}^2\times \frac{1}{0.2}=2.25$
$\therefore$ Ball will first hit the 3rd step first.