Problem source

I am standing next to an ice-cream van at a distance $d$ from the top of a vertical cliff of height $h$. It is not safe for me to go any nearer to the top of the cliff. My niece Padma is on the broad level beach at the foot of the cliff. I have just discovered that I have left my wallet with her, so I cannot buy her an ice-cream unless she can throw the wallet up to me. She can throw it at speed $V$, at any angle she chooses and from anywhere on the beach. Air resistance is negligible; so is Padma's height compared to that of the cliff. Show that she can throw the wallet to me if and only if 
\[
V^{2}\geqslant g(2h+d).
\]

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2-(#1)^2};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-7};
    \def\yu{3};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        \draw[thick, blue, smooth, domain={-sqrt(2)}:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

        \draw (\xl, {0}) -- ({sqrt(2)}, {0}) -- ({sqrt(2)}, \yl); 

        \draw[<->] ({-sqrt(2)}, {-.2}) -- ({sqrt(2)}, {-.2}) node[pos=0.5, below] {$d$};
        \draw[<->] ({-0.2 +sqrt(2)}, {0}) -- ({-0.2 + sqrt(2)}, \yl) node[pos=0.5, left] {$h$};

        \filldraw ({sqrt(2)}, {0}) circle (1.5pt) node[above right] {$(0,0)$};

        \draw[<->] ({sqrt(2)}, {\yl+.2}) -- ({sqrt(2-(\yl+.2))}, {\yl+.2}) node[pos=0.5, above] {$x_0$};
        

    \end{scope}
    \end{tikzpicture}
\end{center}

Rather than considering Padma's throw, imagine a throw in reverse from me. As we can see from the diagram, it will need to pass through $(0,0)$ to have minimal speed when it hits the ground, so possible throws are:

\begin{align*}
    && 0 &= u \sin \alpha t - \frac12 g t^2 \\
    \Rightarrow && T &= \frac{2u \sin \alpha}{g} \\
    && d &= u \cos \alpha T \\
    \Rightarrow && \frac{d}{u \cos \alpha} &=  \frac{2u \sin \alpha}{g} \\
    \Rightarrow && dg &= u^2 \sin 2 \alpha \\
    && v^2 &= u^2 + 2as \\
    \Rightarrow && V_y^2 &= u^2 \sin^2 \alpha + 2gh \\
    \Rightarrow && V^2 &=  u^2 \sin^2 \alpha + 2gh + u^2 \cos^2 \theta \\
    &&&= u^2 + 2gh \\
    &&&= 2gh + \frac{dg}{\sin 2 \alpha} \geq 2gh +dg = g(2h+d)
\end{align*}