Analysis

We consider a particle projected horizontally with velocity $u$ from the origin $(0,0)$. We define the coordinate system with the $x$-axis horizontal and the $y$-axis vertically downwards (along gravity).

At any time $t$:

• Position Vector ($\vec{r}$): $\vec{r} = ut \hat{i} + \frac{1}{2}gt^2 \hat{j}$
• Velocity Vector ($\vec{v}$): $\vec{v} = u \hat{i} + gt \hat{j}$
• Acceleration Vector ($\vec{a}$): $\vec{a} = g \hat{j}$ (Acceleration due to gravity is constant)

(a) & (b) Velocity Components

Radial Velocity ($v_r$): The component of velocity along $\vec{r}$. $$ v_r = \frac{\vec{v} \cdot \vec{r}}{|\vec{r}|} = \frac{(u)(ut) + (gt)(\frac{1}{2}gt^2)}{r} = \frac{u^2t + \frac{1}{2}g^2t^3}{\sqrt{u^2t^2 + \frac{1}{4}g^2t^4}} $$ Simplifying by dividing numerator and denominator by $t$: $$ v_r = \frac{u^2 + \frac{1}{2}g^2t^2}{\sqrt{u^2 + \frac{1}{4}g^2t^2}} $$ Both numerator and denominator increase with $t$, but the numerator grows faster (factor of $1/2$ vs $1/4$ under root). Mathematically, as $t \to \infty$, $v_r \to gt$. It starts at 0 and always increases.

Transverse Velocity ($v_\theta$): The component perpendicular to $\vec{r}$. $$ v_\theta = \frac{|\vec{v} \times \vec{r}|}{|\vec{r}|} = \frac{|(u)(\frac{1}{2}gt^2) – (gt)(ut)|}{r} = \frac{\frac{1}{2}ugt^2}{\sqrt{u^2t^2 + \frac{1}{4}g^2t^4}} $$ Dividing by $t^2$: $$ v_\theta = \frac{\frac{1}{2}ug}{\sqrt{\frac{u^2}{t^2} + \frac{1}{4}g^2}} $$ As $t$ increases, the term $\frac{u^2}{t^2}$ decreases, causing the denominator to decrease. Thus, the whole fraction always increases (approaching a limit of $u$).

(c) & (d) Acceleration Components

Radial Acceleration ($a_r$): Component of $\vec{g}$ along $\vec{r}$. Let $\phi$ be the angle between $\vec{r}$ and the vertical $y$-axis. Then $a_r = g \cos\phi$. $$ \cos\phi = \frac{y}{r} = \frac{\frac{1}{2}gt^2}{\sqrt{u^2t^2 + \frac{1}{4}g^2t^4}} = \frac{\frac{1}{2}gt}{\sqrt{u^2 + \frac{1}{4}g^2t^2}} $$ This function starts at 0 (when $t=0$) and approaches 1 (when $t \to \infty$). Thus, $a_r$ always increases.

Transverse Acceleration ($a_\theta$): Component of $\vec{g}$ perpendicular to $\vec{r}$. $$ a_\theta = g \sin\phi $$ Since the total acceleration magnitude is constant ($g$), and $a_r$ is increasing, $a_\theta = \sqrt{g^2 – a_r^2}$ must always decrease.