Question 14(d) | BetaMath

Question 14(d)

A particle is projected from the origin, with initial speed $V$ at an angle of $\theta$ to the horizontal. The position vector of the particle, $\underset{\sim}{r}(t)$ , where $t$ is the time after projection and $g$ is the acceleration due to gravity, is given by $\underset{\sim}{r}(t) = \begin{pmatrix}Vt\cos\theta\\Vt\sin\theta-\frac{gt^2}2\end{pmatrix}$

Let $D(t)$ be the distance of the particle from the origin at time $t$ , so $D(t) = |\underset{\sim}{r}(t)|$ . Show that for $\displaystyle \theta < \sin^{-1}\sqrt\frac89$ the distance, $D(t)$ , is increasing for all $t > 0$ .

Solution

We compute the distance of the particle at time $t$ by taking the magnitude of the position vector

\begin{aligned} D(t) = |\underset{\sim}{r}(t)| \\ D(t) = \left|\begin{pmatrix}Vt\cos\theta\\Vt\sin\theta-\frac{gt^2}2\end{pmatrix}\right| \\ D(t) = \sqrt{\left(Vt\cos\theta\right)^2 + \left(Vt\sin\theta-\frac{gt^2}2\right)^2} \\ D(t) = \sqrt{V^2t^2\cos^2\theta + V^2t^2\sin^2\theta -2 \left(Vt\sin\theta\right)\frac{gt^2}2+\left(\frac{gt^2}2\right)^2} \\ D(t) = \sqrt{V^2t^2 - gt^2\left(Vt\sin\theta\right)+\left(\frac{gt^2}2\right)^2} \\ D(t) = \sqrt{V^2t^2 - gVt^3\sin\theta + \frac{g^2t^4}4} \\ \end{aligned}

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