A body is thrown upward at intial velocity of `7 ms^(-1)` , then at which height it's kinetic energy becomes half ?

Explanation

Suppose at height h its kinetic becomes halve to its initial kinetic energy .
`K = 1/2 (K_(0))/2 `
`1/2 mv^(2) = 1/2 (1/2mv_(0)^(2))`
` :. V^(2) = 1/2 v_(0)^(2) = 2gh" "…(1)`
but equation of a body thrown in upward direction .
`v_(2)^(2) -v_(0)^(3) =2gh`
`1/2v_(0)^(2) =v_(0)^(2) = - 2gh ` [ From equation (1)]
` :. -1/2 v_(0)^(2) = - 2gh`
` :. 2gh = 1/2 v_(0)^(2)`
` :. h = (v_(0)^(2))/(4g) = 49/(4xx9.8)`
` :. h = 1.25 m `
Alternative Method :
From conservation of mechanical energy
`E = K+V`
` :. ` Constant ` = K+V`
For at point on ground and at point on height h
`K_(1)+V_(1)=K_(2)+V_(2)`
` :. K_(1) + 0 = (K_(1))/2 +V_(2)`
` :. V_(2) = K_(1) - (K_(1))/2 `
` :. V_(2) = (k_(1))/2`
` :. "mgh " = 1/2 (1/2 mv_(0)^(2))`
` :. h = (v_(0)^(2))/(4g)`
`((7)^(2))/(4xx9.8)`
` :. h = 1.25 m `