A ball is projected vertically upwards from ground with velocity $v=24 \mathrm{m} {\mathrm{s}}^{-1}$. Assume the air resistance produces a constant retardation of $2 \mathrm{m} {\mathrm{s}}^{-2}$. Then $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

A particle starts moving along$\mathrm{x}$-axis from origin with initial velocity of $3 \mathrm{m} {\mathrm{s}}^{-1}$. The acceleration a of the particle varies with position $\mathrm{x}$ as $\mathrm{a}={\mathrm{Kx}}^2$, where $\mathrm{K}$ is a constant. At $\mathrm{x}=\sqrt{3} \mathrm{m}$, the tangent to the graph makes an angle of $60°$ with $\mathrm{x}$-axis. Then at $\mathrm{x}=\sqrt{3} \mathrm{m}$ ( $\mathrm{v}$ represents velocity of particle at position $\mathrm{x}$ )

For the $v-t$ graph shown in figure match the following columns.