JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Kinematics

During the fall from height $h$ the ball starts from rest, so $v^2=2g(h-y)$, where $y$ is the height above the floor; this is a sideways parabola in the $v$--$y$ plane, with $v=0$ at $y=h$ and the speed greatest, $v=\sqrt{2gh}$, at the floor $y=0$. After the bounce the ball rises only to $h/2$, so on the way up it follows $v^2=2gy$ but reaches zero at $y=h/2$, and its speed leaving the floor is smaller than the impact speed. The graph showing a parabolic speed--height curve that peaks at the floor, falls to zero at $y=h$ on the way down and to zero at $y=h/2$ on the way up represents the motion.