JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Kinematics

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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 the step 5 of the stairway will be

x

^{- 1}

where

x =

______ [use

g = 10

m/s

^{2}

Answer: 2

Solution

Each step is

0.1 m

high and

0.1 m

wide. For the ball to just clear steps 1-4 and strike the edge of step 5, it must fall through 4 step-heights (

0.4 m

) while covering 4 step-widths (

0.4 m

) horizontally.

Vertical:

0.4 = \frac{1}{2} g t^{2} = \frac{1}{2} (10) t^{2} ⟹ t^{2} = 0.08 s^{2}

Horizontal:

0.4 = u t ⟹ u = \frac{0.4}{t}

u^{2} = \frac{0.16}{t ^{2}} = \frac{0.16}{0.08} = 2 ⟹ u = 2 m/s

x = 2

.

Key confirmed: 2.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer

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