A body of mass $1kg$ thrown upwards with a velocity of $10m/s$ comes to rest (momentarily) after moving up by $4m$. The work done by air drag in this process is (Take $g=10m/{ s }^{ 2 }$)

$\begin{array}{l}n=1 \mathrm{~m} \\ h=4 \mathrm{~m} \\ v=10 \mathrm{~m} / \mathrm{s} \\ g=10 \mathrm{~kg}~m / \mathrm{s}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{2} m v^{2}+\omega=m g h \\ \Rightarrow \omega=m g h-\frac{1}{2} n v^{2} \\ =1 \times 10 \times 4-\frac{1}{2} \times 1 \times 10^{2} \\ =40-50 \\ =-10 \mathrm{~J}\end{array}$