A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits step 5 of the stairway will be , where = _____.

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Solution

Q.
A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits step 5 of the stairway will be $\sqrt{x} m s^{- 1}$ , where $x$ = _____.

[Use g = 10 $m / s^{2}$ ]. [2024]

Ans.
(2)

The ball needs to just cross 4 steps to just hit the 5th step.

Therefore, horizontal range (R) = 0.4 m.

$R = u \cdot t$

In vertical direction, $h = \frac{1}{2} g t^{2}$

$0.4 = \frac{1}{2} g t^{2} = \frac{1}{2} g {(\frac{0.4}{u})}^{2}$

$u^{2} = 2 \Rightarrow u = \sqrt{2} m/s$

Therefore, $x = 2$

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