A small child tries to mimic his father by lifting a mass of 10 kg on his head. As soon as he succeeds in lifting it, he loses the object that falls back to the ground. If the child has a height of 90cm, then the kinetic energy at half the height of the child is:

Given, 

Mass lifted by the child (m) = 10 kg
Height of a child (h₁) = 90 cm = 0.9 m  

Half of the height (h₂) = 0.45 m
g = 9.8 m/s²
According to the Law of conservation of energy, the total energy remains constant.
So, the initial total energy $=$ total energy at half of the height
$\mathrm{ }(\mathrm{K}.\mathrm{E}.+\mathrm{P}.\mathrm{E}.{)}_{\mathrm{initial}\mathrm{ }}=(\mathrm{K}.\mathrm{E}+\mathrm{P}.\mathrm{E}.{)}_{\mathrm{half} \mathrm{the} \mathrm{height}}$
$\Rightarrow \mathrm{ }(\frac{1}{2} {\mathrm{mv}}^2+{\mathrm{mgh}}_1)=(\mathrm{K}.\mathrm{E}.+{\mathrm{mgh}}_2)$ 

$\Rightarrow$$(0+{\mathrm{mgh}}_1)=( \mathrm{K}.\mathrm{E}.+{\mathrm{mgh}}_2)$    [Since, the initial velocity of a body v = 0]
$\Rightarrow \mathrm{K}.\mathrm{E}.={\mathrm{mgh}}_1-{\mathrm{mgh}}_2$

$\Rightarrow \mathrm{K}.\mathrm{E}.=\mathrm{mg}({\mathrm{h}}_1-{\mathrm{h}}_2)$
$\Rightarrow \mathrm{K}.\mathrm{E}.=10\times 9.8\mathrm{ }(0.9-0.45)$
$\Rightarrow$K.E. = 44.1J
 Therefore, If the child has a height of 90cm, then the kinetic energy at half the height of the child is 44.1J.