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A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_(0) It collides with the ground , loses 50 percent ofits energyin collision and reboundsto the same height .the initial velocity v_(0) is : (Take g = 10 ms^(-2))

Answer» <html><body><p>`10ms^(-1)`<br/>`14 ms^(-1)`<br/>`20ms^(-1)`<br/>`28ms^(-1)`</p>Solution :Let ball rebounds with speed v so<br/> ` :. "In" v^(2) -v_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)^(2) =2gh ,v_(0) = 0 ` <br/> ` :. V= sqrt(2gh) =sqrt(2xx10xx20) = 20 m//s `<br/> <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> just after rebound <br/> `E =1/2 mv^(2) =1/2 m xx m xx (20)^(2)` <br/> ` = 1/2 m xx400` <br/> = <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> m Joule<br/> 50 % energy <a href="https://interviewquestions.tuteehub.com/tag/loses-1079374" style="font-weight:bold;" target="_blank" title="Click to know more about LOSES">LOSES</a> in <a href="https://interviewquestions.tuteehub.com/tag/collision-922060" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISION">COLLISION</a> means just before collisionenergyis 400 m <br/> By using energy conservation <br/> `1/2 mv_(0)^(2) +mgh = 400 m `<br/> ` :. v_(0)^(2) = 800 - 2gh`<br/> ` :. v_(0)^(2) = 800 - 400` <br/> ` :. v_(0)^(2) = 400 `<br/> ` :. v_(0) = 20 ms^(-1)`</body></html>


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