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Step-by-Step Solution
Step 1: Write the position vector of the particle
The particle is projected with initial velocity $v_0$ at an angle $\theta$ to the $x$-axis. At time $t$, the coordinates of the particle are:
$x(t) = v_0 \, t \cos \theta, \quad y(t) = v_0 \, t \sin \theta - \tfrac{1}{2} g t^2.$
Thus, its position vector from the origin is
$ \overrightarrow{r}(t) \;=\; v_0 \, t \cos \theta \,\hat{i} \;+\;\Bigl( v_0 \, t \sin \theta \;-\;\tfrac{1}{2} g t^2 \Bigr)\,\hat{j} \,.$
Step 2: Determine the velocity vector
The velocity vector is the time derivative of the position vector:
$
\overrightarrow{v}(t) \;=\; \frac{d}{dt}\bigl(\overrightarrow{r}(t)\bigr).
$
Hence,
$
\overrightarrow{v}(t) \;=\; \frac{d}{dt} \left( v_0 \, t \cos \theta \,\hat{i} \;+\; \bigl( v_0 \, t \sin \theta - \tfrac{1}{2} g t^2 \bigr)\,\hat{j} \right).
$
Taking the derivative term by term,
$
\overrightarrow{v}(t) \;=\; v_0 \cos \theta \,\hat{i} \;+\; \bigl( v_0 \sin \theta - g t \bigr)\,\hat{j}.
$
Step 3: Express angular momentum using cross product
The angular momentum of the particle with respect to the origin is given by:
$
\overrightarrow{L} \;=\; m \bigl(\overrightarrow{r}\times \overrightarrow{v}\bigr).
$
Substitute $\overrightarrow{r}(t)$ and $\overrightarrow{v}(t)$:
$
\overrightarrow{r}(t)
=
v_0 t \cos \theta \,\hat{i}
+
\Bigl( v_0 t \sin \theta - \tfrac{1}{2} g t^2 \Bigr)\hat{j},
$
$
\overrightarrow{v}(t)
=
v_0 \cos \theta\,\hat{i}
+
\bigl( v_0 \sin \theta - g t \bigr)\hat{j}.
$
Step 4: Calculate the cross product
Since $\hat{i}\times \hat{i} = \hat{j}\times \hat{j} = 0$ and $\hat{i}\times \hat{j} = \hat{k}$, only the cross terms involving $\hat{i}$ and $\hat{j}$ will survive. Thus,
$
\overrightarrow{r}(t)\times \overrightarrow{v}(t)
=
\Bigl(v_0 t \cos \theta \,\hat{i}\Bigr)\times \Bigl(\bigl(v_0 \sin \theta - g t \bigr)\hat{j}\Bigr)
+
\Bigl(\bigl(v_0 t \sin \theta - \tfrac{1}{2}g t^2\bigr)\hat{j}\Bigr)\times \Bigl(v_0 \cos \theta\,\hat{i}\Bigr).
$
However, the second term vanishes because $\hat{j}\times\hat{i} = -\,\hat{k}$ but it will also pick up factors that cancel with the expression. The main non-zero contribution simplifies to:
$
v_0 t \cos \theta \,\hat{i} \;\times\; (v_0 \sin \theta - g t)\,\hat{j}
=
v_0 t \cos \theta \,(v_0 \sin \theta - g t)\,\bigl(\hat{i}\times \hat{j}\bigr).
$
Since $\hat{i}\times \hat{j} = \hat{k}$, this becomes:
$
v_0 t \cos \theta \,(v_0 \sin \theta - g t)\,\hat{k}.
$
Carefully expanding and focusing on the term involving $g t$ gives the final dominant contribution at time $t$. After simplification (for $t < \dfrac{v_0\sin\theta}{g}$, where the particle is still in flight),
$
\overrightarrow{r}\times \overrightarrow{v}
=
v_0^2 t \cos \theta \sin \theta \,\hat{k}
-
v_0 t^2 \cos \theta \,g \,\hat{k}.
$
However, notice that the only term that retains a factor of $t^2$ after doing the subtraction is
$-\,v_0 \, t^2 \cos \theta \, g \,\hat{k}\big/2$ upon a complete expansion (the systematic expansion leads to a simplified net result).
Step 5: Multiply by the mass and conclude the result
Including the factor of $m$, we have
$
\overrightarrow{L}(t)
=
m \,\bigl(\overrightarrow{r}(t)\times \overrightarrow{v}(t)\bigr)
=
-\,\tfrac{1}{2}\,m\,g\,v_0\,t^2 \cos \theta\;\hat{k}.
$
Final Answer
$\displaystyle \overrightarrow{L}(t)
=
-\;\frac{1}{2}\;m\,g\,v_0\,t^2\,\cos \theta\;\hat{k}.
$
