A ball is thrown vertically up (taken as +z- | vedclass

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(A) The momentum of the ball is given by $p = mv$,where $m$ is the mass and $v$ is the velocity.For motion under gravity,the relationship between velo

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(A) The momentum of the ball is given by $p = mv$,where $m$ is the mass and $v$ is the velocity.For motion under gravity,the relationship between velocity $v$ and height $h$ is given by $v^2 = u^2 - 2gh$,where $u$ is the initial velocity.Substituting $v = p/m$ into the equation,we get $(p/m)^2 = u^2 - 2gh$.Rearranging this,we get $p^2 = m^2u^2 - 2gm^2h$,which can be written as $p^2 = -2gm^2h + m^2u^2$.This is the equation of a parabola of the form $p^2 = -Ah + B$,where $A = 2gm^2$ and $B = m^2u^2$.As the ball moves up,$p$ decreases from $mu$ to $0$ as $h$ increases from $0$ to $H_{max}$. As it falls back down,$p$ becomes negative and increases in magnitude from $0$ to $-mu$ as $h$ decreases from $H_{max}$ to $0$.The correct graph is a parabola opening towards the negative $h$-axis,starting at $p = mu$ (positive),passing through $p=0$ at $h=H_{max}$,and ending at $p = -mu$ (negative) at $h=0$.

$A$ ball is thrown vertically up (taken as $+z-$ axis) from the ground. The correct momentum-height $(p-h)$ diagram is

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