Question

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after being released?

Answer 1

Potential energy at any point is (M G H). On the way down, only H changes. So halfway down, half of the potential energy remains, and the other half has turned to kinetic energy. Half of the (M G H) it had at the tpp is (0.5 x 9.8 x 10) = 49 joules.

Answer 2

Answer: The kinetic energy of the ball when it reaches halfway after being released is 49 Joules.

Step-by-step explanation:

Sum of potential energy and kinetic energy always remains constant while freely falling of the body.

At initial point : (when ball is at verge to fall down)

P.E+K.E=
`mgh+(1)/(2)mv^2=0.50* 9.8m/s^2* 20 m+(1)/(2)* 0.50 kg* (0 m/s)^2=98 Joules`

At the point when ball reaches half way at height of h', h'=10 m

P.E+K.E=

`=mgh'+(1)/(2)mv'^2=0.50* 9.8m/s^2* 10 m+(1)/(2)* 0.50 kg* (v' m/s)^2=49 Joules+(1)/(2)* 0.50 kg* (v' m/s)^2`

`P.E+K.E=98 J=49 Joules+(1)/(2)* 0.50 kg* (v' m/s)^2`

`K.E=(1)/(2)mv'^2=98 J-49 J=49 J`

The kinetic energy of the ball when it reaches halfway after being released is 49 Joules.