Question 27: Motion in 2D & Vectors

Question

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the $\mathrm{x}-$ axis with an initial velocity $\mathrm{v}_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

Accepted Answer

$$ \vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}. $$
At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:
$$ \frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}. $$
The coordinates at this time are:
$$ x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g}, $$
$$ y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}. $$
Thus, the position vector at maximum height is:
$$ \vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}. $$
At maximum height, the only nonzero component of velocity is the horizontal one:
$$ \vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}. $$
The angular momentum about the origin is given by:
$$ \vec{L} = m\,\vec{r} 	imes \vec{v}. $$
Writing out the cross product:
$$ \vec{r} 	imes \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}ight) 	imes \frac{v_0}{\sqrt{2}}\,\hat{i}. $$
Since the cross product of $\hat{i}$ with itself is zero and:
$$ \hat{j} 	imes \hat{i} = -\hat{k}, $$
we have:
$$ \vec{r} 	imes \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}. $$
Thus, the angular momentum is:
$$ \vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}. $$
This means the magnitude is:
$$ \frac{m\,v_0^3}{4\sqrt{2}g}, $$
and its direction is along the negative $$\hat{k}$$ (or negative $$z$$-axis).

Answer

$\frac{m v_o{ }^3}{2 \sqrt{2} g}$ along negative $z$-axis

Answer

$\frac{m v_o^3}{2 \sqrt{2} g}$ along positive $z$-axis

Answer

$\frac{m v_o^3}{4 \sqrt{2} g}$ along positive $z$-axis

Answer

$\frac{m v_o^3}{4 \sqrt{2} g}$ along negative z-axis

Motion in 2D & Vectors

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