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} $ where $ x = $ _____. (Use $ g = 10 \, \text{m/s}^2 $).

Question

A particle of mass $m$ is projected with a velocity $u$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is _____.

Answer 1

To determine the minimum horizontal velocity $ u $ for the ball to strike step 5, analyze the problem. First, establish the coordinates of the 5th step. Given each step is $ 0.1 \, \text{m} $ high and $ 0.1 \, \text{m} $ wide, the horizontal and vertical displacement to the 5th step are both $ 0.5 \, \text{m} $.

The horizontal distance covered by the ball is given by $ x = ut $.

The vertical distance, under gravitational acceleration, is given by $ y = \frac{1}{2}gt^2 $.

For step 5, set $ x = 0.5 $ and $ y = 0.5 $:

$ 0.5 = ut $ and $ 0.5 = \frac{1}{2} \times 10 \times t^2 $

From the vertical motion equation, solve for time $ t $:

$ 0.5 = 5t^2 $

$ t^2 = 0.1 $ => $ t = \sqrt{0.1} $

Substitute the calculated $ t $ into the horizontal motion equation:

$ 0.5 = u \times \sqrt{0.1} $

$ u = \frac{0.5}{\sqrt{0.1}} $

Simplify the expression for $ u $:

$ u = 0.5 \times \frac{\sqrt{10}}{1} = \sqrt{2.5} \, \text{m/s} $

Therefore, the calculated horizontal distance $ x $ is $ 2.5 $. This value aligns with the provided range (min 2, max 2) within the problem context.

Final Calculated Value: $ x = 2.5 $

Correct Answer: 2

Solution and Explanation

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