First slide

Relative motion in 2 dimension

Question

A swimmer is approaching the shore with a speed of $5 \sqrt{3} {ms}^{- 1}$ . At the instant when it is at a distance of $30 \sqrt{3} m$ from the shore, he projects a ball at an elevation of 30^o for it to just reach the shore. The speed of stone relative to swimmer is 5x ms^-1. The value of x is ________.

Difficult

Solution

$Here, u_{x} = 5 \sqrt{3} + u_{rel} \cos 30^{\circ}$
$and u_{y} = u_{rel} \sin 30^{\circ}$
$∴ 30 \sqrt{3} = (5 \sqrt{3} + u_{rel} \cos 30^{\circ}) T$
$and T = \frac{2 u_{y}}{g} = \frac{2 u_{rel} \sin 30^{\circ}}{g} = \frac{u_{rel}}{g}$
$\begin{array}{l} ∵ 30 \sqrt{3} = \frac{5 \sqrt{3} u_{rel}}{g} + \frac{\sqrt{3}}{2 g} u_{rel}^{2} \\ \Rightarrow 300 = 5 u_{rel} + \frac{u_{rel}^{2}}{2} \end{array}$
$\begin{array}{l} \Rightarrow \frac{u_{rel}^{2}}{2} + 5 u_{rel} - 300 = 0 \\ \Rightarrow u_{rel}^{2} + 10 u_{rel} - 600 = 0 \\ ∴ u_{rel} = \frac{- 10 \pm \sqrt{100 + 2400}}{2} = \frac{- 100 \pm 50}{2} \\ = \frac{40}{2} = 20 {ms}^{- 1} \end{array}$
$∴ n = 4$

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