Q 11-03-068JEE MainJEE Main 2024 (29 Jan, Shift 1)Hard
A ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1\ \text{m}$ high and $0.1\ \text{m}$ wide. The minimum velocity $u$ with which the ball just hits the step 5 of the stairway will be $\sqrt x\ \text{m s}^{-1}$, where $x =$ ______ [use $g = 10\ \text{m s}^{-2}$].
Numerical value type. Enter your answer.
Answer: 2
To land on step 5, the ball must just clear the edge of step 4, which is $0.4\ \text{m}$ below and $0.4\ \text{m}$ ahead of the start:
$$0.4 = \frac12(10)t^2 \;\Rightarrow\; t^2 = 0.08$$
$$u = \frac{0.4}{t} \;\Rightarrow\; u^2 = \frac{0.16}{0.08} = 2$$
So $u = \sqrt2\ \text{m s}^{-1}$ and $x = 2$.
Solution by Sreeraj P, M.Sc Physics