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Question: A man is running with the speed of \(10m{{s}^{-1}}\) and it has been found that the rain is falling ...

A man is running with the speed of 10ms−110m{{s}^{-1}} and it has been found that the rain is falling vertically with speed of 20ms−120m{{s}^{-1}}. What speed should the man be having when he is running, such that the rain appears to him at an angle of 37∘37{}^\circ with the vertical?

Explanation

Solution

The velocity of the man will be equivalent to the product of the velocity of the rain and sine of the angle. Substitute the values of the speed of the man and speed of the rain in this in order to get the angle at this situation. This all will help you in answering this question.

Complete answer:
It has been given that the speed of the man who is running can be written as,
Vm=10ms−1{{V}_{m}}=10m{{s}^{-1}}
The speed of the rain falling vertically down can be written as,
Vr=20ms−1{{V}_{r}}=20m{{s}^{-1}}
The angle between them is to be,
θ=37∘\theta =37{}^\circ
As we can see the velocity of the man will be equivalent to the product of the velocity of the rain and sine of the angle. That is we can write that,
Vrsin⁡θ=Vm{{V}_{r}}\sin \theta ={{V}_{m}}
Substituting the values of the speed of the man and speed of the rain in this in order to get the angle at this situation.
That is,
20×sin⁡θ=10 θ=sin⁡−112=30∘ \begin{aligned} & 20\times \sin \theta =10 \\\ & \theta ={{\sin }^{-1}}\dfrac{1}{2}=30{}^\circ \\\ \end{aligned}
Now we can write that,
V⃗m=V⃗{{\vec{V}}_{m}}=\vec{V}
That is,
Vrcos⁡30∘=R−1cos⁡37∘{{V}_{r}}\cos 30{}^\circ ={{R}^{-1}}\cos 37{}^\circ
Where RR be the resultant. Substituting the values in it will give,
20×32=R×45 ⇒R=2532 \begin{aligned} & 20\times \dfrac{\sqrt{3}}{2}=R\times \dfrac{4}{5} \\\ & \Rightarrow R=\dfrac{25\sqrt{3}}{2} \\\ \end{aligned}
The resultant has been obtained.
Now we can write that,
Vrm=Vrsin⁡30∘−Rsin⁡37∘{{V}_{rm}}={{V}_{r}}\sin 30{}^\circ -R\sin 37{}^\circ
Therefore the velocity of the man with respect to the rain can be written as,
Vrm=20×12−2533×35 ⇒Vrm=10−1532 ∴Vrm=532 \begin{aligned} & {{V}_{rm}}=20\times \dfrac{1}{2}-\dfrac{25\sqrt{3}}{3}\times \dfrac{3}{5} \\\ & \Rightarrow {{V}_{rm}}=10-\dfrac{15\sqrt{3}}{2} \\\ & \therefore {{V}_{rm}}=\dfrac{5\sqrt{3}}{2} \\\ \end{aligned}
Therefore the answer has been calculated.

Note:
The relative velocity is defined as the velocity of a body or an observer B which is being placed in a rest frame of any other body or observer A. The relative velocity will differ with respect to the frame of reference the observer chooses to observe. The relative velocity can be found by observing from a moving frame also.