Equation of trajectory of a projectile is given as $y=x-\frac{x^2}{40}.$ The radius of curvature at highest point of its trajectory is [Take, $g=10 \mathrm{m} {\mathrm{s}}^{-2}$ ]

A ball is thrown with velocity $\overrightarrow{u}=(3\hat{i}+4\hat{j})\mathrm{m} {\mathrm{s}}^{-1}$ towards a wall, where it hits at height $h_1$ at time $t_1$ after the launch.

Another ball is projected from same point with velocity $\overrightarrow{u}=(3\hat{i}+5\hat{j})\mathrm{m} {\mathrm{s}}^{-1}$ towards the wall where it hits at height $h_2$ at time $t_2$ after the launch. Then

Two particles $A$ and $B$ are at rest at separation of $160 \mathrm{m}.$ Both the particle starts moving at $t=$ $0.$ Particle A moves in a straight line perpendicular to initial position $AB$ with constant acceleration of $3 \mathrm{m} {\mathrm{s}}^{-2}.$ Particle $B$ moves in such a way that it always aims towards $A$ and moves with speed increasing at constant rate of $5 \mathrm{m} {\mathrm{s}}^{-2}.$ Then

A particle is projected from ground to ground and its path is as shown in the figure. Points $A$ and $B$ are at same horizontal level and points $C$ and $D$ are at another same horizontal level. Separation between line $AB$ and $CD$ is $5 \mathrm{m}.$ If particle takes time $t_1$ to go from $A$ to $B$ and time $t_2$ from $C$ to $D,$ then is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$