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))

A ball is thrown vertically downwards from a height of 20 m with an in

1.	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» `10ms^(-1)` `14 ms^(-1)` `20ms^(-1)` `28ms^(-1)` Solution :Let ball rebounds with speed v so ` :. "In" v^(2) -v_(0)^(2) =2gh ,v_(0) = 0 ` ` :. V= sqrt(2gh) =sqrt(2xx10xx20) = 20 m//s ` ENERGY just after rebound `E =1/2 mv^(2) =1/2 m xx m xx (20)^(2)` ` = 1/2 m xx400` = 200 m Joule 50 % energy LOSES in COLLISION means just before collisionenergyis 400 m By using energy conservation `1/2 mv_(0)^(2) +mgh = 400 m ` ` :. v_(0)^(2) = 800 - 2gh` ` :. v_(0)^(2) = 800 - 400` ` :. v_(0)^(2) = 400 ` ` :. v_(0) = 20 ms^(-1)`