A body of mass 1 kg is thrown with a velocity of `250 cm *s^(-1)`, from a height 8m above the earth's surface . What will be its kinetic energy just before it touches the ground ?

Explanation

Let the speed of the body just before it touches the ground be v .
Taking the upward direction as positive ,
`v^2 =u^2+2gh =(2.5)^2+2xx9.8xx8 =163.05 m^2 *s^(-2)`
[ `because u=250 cm*s^(-1) =2.5 m*s^(-1), g=-9.8 m*s^(-2) and h= -8 m]`
`therefore` Kinetic energy just before it touches the ground
`=1/2 mv^2 =1/2 xx1xx163.05 =81525 J`.