3.3 Relative velocity in two dimensions (application)

Problem : A man, moving at 3 km/hr along a straight line, finds that the rain drops are falling at 4 km/hr in vertical direction. Find the angle with which rain drop hits the ground.

Solution : Let the man be moving in x-direction. Let us also denote man with “A” and rain drop with “B”. Here, we need to know the direction of rain drop with respect to ground i.e. the direction of ${\mathbf{v}}_B$ .

Here,

$$\begin{array}{l}{v}_A=3\mathrm{km}\mathrm{/}\mathrm{hr}\\ v_B=?\\ v_{\mathrm{BA}}=4\mathrm{km}\mathrm{/}\mathrm{hr}:\text{in the vertical direction}\end{array}$$

Using equation, ${\mathbf{v}}_{\mathrm{BA}}={\mathbf{v}}_B-{\mathbf{v}}_A$ ,

$$\begin{array}{l}\Rightarrow {\mathbf{v}}_B={\mathbf{v}}_A+{\mathbf{v}}_{\mathrm{BA}}\end{array}$$

In order to evaluate the right hand side of the equation, we construct the vector diagram as shown in the figure.

From inspection of given data and using appropriate trigonometric function in ΔOBR, we have :

$$\begin{array}{l}\tan \theta =\frac{3}{4}=\tan {37}^0\\ \Rightarrow \theta ={37}^0\end{array}$$

Problem : A person, moving at a speed of 1 m/s, finds rain drops falling (from back) at 2 m/s at an angle 30° with the vertical. Find the speed of raindrop (m/s) with which it hits the ground.

Solution : Let the person be moving in x-direction. Let us also denote man with “A” and rain drop with “B”. Here we need to know the speed of the rain drops with respect to ground i.e. ${\mathbf{v}}_B$ .

Here,

$$\begin{array}{l}{v}_A=1m\mathrm{/}s\\ v_B=?\\ v_{\mathrm{BA}}=2m\mathrm{/}s\end{array}$$

Using equation, ${\mathbf{v}}_{\mathrm{BA}}={\mathbf{v}}_B-{\mathbf{v}}_A$ ,

$$\begin{array}{l}\Rightarrow {\mathbf{v}}_B={\mathbf{v}}_A+{\mathbf{v}}_{\mathrm{BA}}\end{array}$$

In order to evaluate the right hand side of the equation, we construct the vector diagram as shown in the figure.

From parallelogram theorem,

$$\begin{array}{l}{v}_B=\surd ({v_A}^2+{v_{\mathrm{BA}}}^2+2v_A{v}_{\mathrm{BA}}\cos {60}^0)\\ \Rightarrow v_B=\surd (1^2+2^2+2x1x2x\frac{1}{2})\\ \Rightarrow v_B=\surd (1+4+2)=\surd 7m\mathrm{/}s\end{array}$$

Problem : A boy moves with a velocity 0.5 i – j in m/s. He receives rains at a velocity 0.5 i – 2 j in m/s. Find the speed at which rain drops meet the ground.

Solution : Let the person be moving along OA. Let us also denote man with “A” and rain drop with “B”. Here we need to know the speed at which rain drops fall on the ground ( $v_B$ ).

Here,

$$\begin{array}{l}{\mathbf{v}}_A=0.5\mathbf{i}-\mathbf{j}\\ v_B=?\\ {\mathbf{v}}_{\mathrm{BA}}=0.5\mathbf{i}-2\mathbf{j}\end{array}$$

Using equation, ${\mathbf{v}}_{\mathrm{BA}}={\mathbf{v}}_B-{\mathbf{v}}_A$ ,

$$\begin{array}{l}\Rightarrow {\mathbf{v}}_B={\mathbf{v}}_A+{\mathbf{v}}_{\mathrm{BA}}\end{array}$$

$$\begin{array}{l}\Rightarrow {\mathbf{v}}_B=0.5\mathbf{i}-\mathbf{j}+0.5\mathbf{i}-2\mathbf{j}=\mathbf{i}-3\mathbf{j}\\ \Rightarrow {\mathbf{v}}_B=\surd (1+9)=\surd \left(10\right)m\mathrm{/}s\end{array}$$