1. Set up the motion equation

A ball is thrown vertically upward from the ground along the $+z$-axis.

Let:

• initial speed $= u$
• height above ground $= h$
• instantaneous speed at height $h$ be $v$
• mass of ball $= m$

Under gravity, $v^2 = u^2 - 2gh$

2. Relate momentum to height

Momentum is $p = mv$

But since the ball can move both upward and downward:

• while going up, $v>0 \Rightarrow p = +m\sqrt{u^2-2gh}$
• while coming down, $v<0 \Rightarrow p = -m\sqrt{u^2-2gh}$

So, $p = \pm m\sqrt{u^2-2gh}$

3. Important features of the $p$-$h$ graph

From $p^2 = m^2(u^2-2gh)$ $p^2 = m^2u^2 - 2m^2gh$

Thus $p$ is not linearly related to $h$; instead, for each height $h$ there are generally two values of momentum:

• one positive (upward journey)
• one negative (downward journey)

Also:

• at $h=0$, $p=+mu$ initially and later $p=-mu$ when it returns to ground
• at maximum height $h_{\max}=\dfrac{u^2}{2g}$, we get $p=0$

So the graph must:

• be symmetric about the $p=0$ axis,
• start from $+mu$ and $-mu$ at $h=0$,
• meet at $p=0$ when $h=h_{\max}$,
• look like two curved branches, not straight lines.

4. Shape of the graph

Writing height in terms of momentum, $h = \frac{u^2}{2g} - \frac{p^2}{2m^2g}$

This is a parabola opening downward in the $(p,h)$ plane.

Hence the correct $p$-$h$ diagram is the one showing two branches: $p = \pm m\sqrt{u^2-2gh}$ with $p=0$ at maximum height.

5. Conclusion

Therefore, the correct option is D.