Bouncy Bouncing Ball!

From a height h=500 mh = 500 \text{ m}, a uniform solid disc of radius R=90 cmR = 90 \text{ cm} is given an angular velocity ω=50003 rad/s\omega = \dfrac{5000}{3} \text{ rad/s} and dropped. The ground is rough and has coefficient of friction μ=1\mu = 1. After each collision, the disc's vertical speed becomes half(50%)(50\%) of the value just before collision. If when the vertical velocity just becomes zero, the horizontal distance travelled by disc is r(in m)r (\text{in m}) and the angular velocity of disc is ω(in rad/s)\omega' (\text{in rad}/\text{s}), enter answer as r+ωr + \omega'.

Take acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2.


Inspiration Aniket Sanghi

All of my problems are original.

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