JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Kinematics

Throwing the ball up, energy conservation gives $v^2=u^2-2gh$, and writing $v=p/m$ this becomes $h=\frac{u^2}{2g}-\frac{p^2}{2g{m}^2}$. So $h$ is a downward-opening parabola in $p$: the height is greatest where $p=0$ (the top of the flight) and falls to zero where $|p|$ is largest (at launch and at landing), and it is symmetric in the sign of $p$. The curve carrying this parabolic, symmetric shape is the correct momentum-height graph.