Question:

A ball rolls off the top of a stairway with horizontal velocity u u . The steps are 0.1m 0.1 \, \text{m} high and 0.1m 0.1 \, \text{m} wide. The minimum velocity u u with which the ball just hits the step 5 of the stairway will be xm/s \sqrt{x} \, \text{m/s} where x= x = _____. (Use g=10m/s2 g = 10 \, \text{m/s}^2 ).

Updated On: Nov 12, 2024
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Correct Answer: 2

Solution and Explanation

Step 1: Determine Horizontal Range to Just Hit Step 5:

- The ball needs to cross 4 steps horizontally to just hit the 5th step. Since each step is 0.1 m wide, the horizontal range R R required to reach the 5th step is:

R=0.4 m R = 0.4 \text{ m}

- Using the horizontal motion equation R=ut R = u \cdot t , we get:

t=Ru=0.4u t = \frac{R}{u} = \frac{0.4}{u}

Step 2: Vertical Motion Analysis:

- For vertical displacement, the ball needs to fall a height of h=4×0.1=0.4 m h = 4 \times 0.1 = 0.4 \text{ m} . Using the vertical motion equation h=12gt2 h = \frac{1}{2} g t^2 :

0.4=1210(0.4u)2 0.4 = \frac{1}{2} \cdot 10 \cdot \left( \frac{0.4}{u} \right)^2

- Simplify to find u u :

0.4=50.16u2 0.4 = 5 \cdot \frac{0.16}{u^2}

u2=2 u^2 = 2

u=2 m/s u = \sqrt{2} \text{ m/s}

Step 3: Determine x x :

- Given that u=x u = \sqrt{x} , we find x=2 x = 2 .

So, the correct answer is: x=2x = 2

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