If we throw a body upwards with velocity of $4 \mathrm{~m} / \mathrm{s}$, at what height does its kinetic energy reduce to half of the initial value? (Take $g=10 \mathrm{~ms}^{-2}$ )

Key Idea : At a given height the half of the kinetic energy of the body is equal to its potential energy. Initial kinetic energy of the body $\begin{aligned} & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} m(4)^2=8 m\end{aligned}$ Let at height $h$, the kinetic energy reduces to half, ie, it becomes 4 m . It is also equal to potential energy. Hence, $m g h=4 \mathrm{~m}$ or $h=\frac{4}{g}=\frac{4}{10}=0.4 \mathrm{~m}$