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.
