Problem source

A particle of mass $m$ is pulled along the floor of a room in a straight line by a light string which is pulled at constant speed $V$ through a hole in the ceiling. The floor is smooth and horizontal, and the height of the room is $h$. Find, in terms of $V$ and $\theta$, the speed  of the particle when the string makes an angle of $\theta$ with the vertical (and the particle is still in contact with the floor). Find also the acceleration, in terms of $V$, $h$ and $\theta$.
Find the tension in the string and hence show that the particle will leave the floor when
\[
\tan^4\theta = \frac{V^2}{gh}\,.
\]

Solution source

\begin{center}
    \begin{tikzpicture}
        \coordinate (A) at (0,0);
        \coordinate (H) at (5, 2);
        \coordinate (E) at (5, 0);
        \draw (A) -- (H);
        \draw (-1, -0.07) -- (6, -0.07);

        \filldraw (A) circle (2pt);
        \draw[dashed] (H) -- ++(0,-2) node[pos=0.5, right] {$h$};
        \pic [draw, angle radius=.8cm, angle eccentricity=1.3, "$\theta$"] {angle = A--H--E};
    \end{tikzpicture}
\end{center}

The length of the string is $h/\cos \theta$, and it is decreasing at a rate $V$. The distance along the ground is decreasing at a rate of $V/\sin \theta$.

Note that $-V = \frac{\d}{\d t} \left ( \frac{h}{\cos \theta} \right) = \frac{h} {\cos^2 \theta} \sin \theta \cdot \dot{\theta} \Rightarrow \dot{\theta} = -\frac{V\cos^2\theta}{h \sin \theta}$.

Note that $a = \frac{\d}{\d t} \left ( \frac{V}{\sin \theta} \right) = -\frac{V}{\sin^2 \theta} \cos \theta \cdot \dot{\theta} = \frac{V^2 \cos^3 \theta}{h\sin^3 \theta}$.

Resolving horizontally we must have $T \sin \theta = ma \Rightarrow T = \frac{V^2m \cos^3 \theta}{h \sin^4 \theta}$.

Resolving vertically at the point where we are about to leave the ground, we must have $T\cos \theta = mg \Rightarrow \frac{V^2m \cos^4 \theta}{h \sin^4 \theta} = mg \Rightarrow \tan^4 \theta = \frac{V^2}{gh}$