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

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

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=$mgh+\frac{1}{2}mv^2=0.50\times 9.8m/s^2\times 20 m+\frac{1}{2}\times 0.50 kg\times (0 m/s)^2=98 Joules$

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

P.E+K.E=

$=mgh'+\frac{1}{2}mv'^2=0.50\times 9.8m/s^2\times 10 m+\frac{1}{2}\times 0.50 kg\times (v' m/s)^2=49 Joules+\frac{1}{2}\times 0.50 kg\times (v' m/s)^2$

$P.E+K.E=98 J=49 Joules+\frac{1}{2}\times 0.50 kg\times (v' m/s)^2$

$K.E=\frac{1}{2}mv'^2=98 J-49 J=49 J$

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

Answer 2

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.