A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. What will be height at which the kinetic energy of the body becomes half of the original value ? (Acceleration due to gravity`=9.8ms^(-2)`)

Explanation

Given, m=5 kg and KE=490 J
From the law of conservation of energy,
`K_(i)+U_(i)=K_(f)+U_(f) impliesK_(i)+0=(K_(i))/(2)+mgh`
`490=245+5xx9.8xxh " " (`:'` K_(f)=(K_(i))/(2))`
`h=(490-245)/(5xx9.8)=(245)/(49)=5 m`