A ball of mass $0.2\text{kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2\text{m}$ while applying the force and the ball goes up to $2\text{m}$ height further, find the magnitude of the force. (Consider $g=10{\text{m/s}}^2$)

Let the force applied by the hand be $F$.The work done by the hand is $W_h=F\times 0.2$.The total displacement from the start of the push to the highest point is $0.2+2=2.2\text{m}$.Applying the Work-Energy Theorem:$W_{net}=\Delta KE$$W_h+W_g=0$ (since it starts and ends at rest)$F(0.2)-mg(2.2)=0$$F(0.2)=0.2\times 10\times 2.2$$F(0.2)=4.4$$F=\frac{4.4}{0.2}=22\text{N}$.Note: The FIITJEE solution provided in the text ($20\text{N}$) likely ignores the $mg$ during the $0.2\text{m}$ push phase ($Fs=mgh⟹F(0.2)=0.2\times 10\times 2⟹F=20$). However, the technically correct answer considering gravity throughout is $22\text{N}$.