JEE Main 2019 — Motion In One Dimension Question with Solution

If the ball was thrown with initial velocity $\mathrm{u}$ , then its velocity at height $\mathrm{\text{'}}{\mathrm{h}}^{\mathrm{\text{'}}}$ can be written as.
$\mathrm{v}=\sqrt{{\mathrm{u}}^2-2\mathrm{g}\mathrm{h}}$
Its linear momentum, $\mathrm{p}=\mathrm{m}\mathrm{v}=\mathrm{m}\sqrt{{\mathrm{u}}^2-2\mathrm{g}\mathrm{h}}$
Or $\mathrm{ }{\mathrm{p}}^2={\mathrm{m}}^2{\mathrm{u}}^2-2{\mathrm{m}}^2\mathrm{g}\mathrm{h}$
The graph between $\mathrm{p}\mathrm{ }\&\mathrm{ }\mathrm{h}$ will be a parabola.
In motion when particle was going up, its velocity is positive and decreasing with height and so is the linear momentum. When it reaches the maximum height, the velocity and linear momentum will be zero. Again, it will start to fall back, the velocity and linear momentum will increase but will have negative value.