Question

Two stones are projected from the top of a tower in opposite directions with the same velocity V at angles 30° and 60° to the horizontal. Then relative velocity of one stone with respect the other is

1) √2v 2) 2v

do with explanation​

Answer 1

Here is the step by step explanation for your question.

Answer 2

Concept: Velocity are often outlined because the rate of amendment of the object’s position with relation to a frame of reference and time. it would sound difficult however speed is largely dashing in an exceedingly specific direction. it's a vector amount, which suggests we want each magnitude (speed) and direction to outline speed.

Find: Calculate the relative velocity of one stone with respect to other stone.

Solution:

V1= Vcosθ1 iˆ+Vsinθ1 jˆ

V2=Vcosθ2(-iˆ) +Vsinθ2 jˆ

V1=V.$\sqrt{3}$/2 iˆ+V/2jˆ

V2=V/2(-iˆ)$\sqrt{3}$/2  jˆ

V1-V2=V.$\sqrt{3}$/2 iˆ+V/2jˆ-(V/2(-iˆ)$\sqrt{3}$/2  jˆ)

        = iˆ($\sqrt{3}$/2+1/2) V+ jˆ(1/2-$\sqrt{3}$/2)V

V12=($\sqrt{3}$+1/2)V iˆ +(1-$\sqrt{3}$/2) V jˆ

V12=$\sqrt{Vx^{2} +Vy^{2} }$  

    =$\sqrt{\frac{\sqrt{3} +1 }{2} ^{2} V^{2} +(\frac{1-\sqrt{3} }{2}) } ^{2}V^{2} }$

    =$\sqrt{\frac{V^{2} }{4} [(\sqrt{3} +1)^{2} +(1-\sqrt{3}) ^{2} }$

    =V/2 $\sqrt{3+1+2\sqrt{3} +1+3-2\sqrt{3} }$

    =V/2 $\sqrt{4+4}$

    =V/2 $\sqrt{8}$

    =V/2 $2\sqrt{2}$

    = $\sqrt{2} V$m/s

Final answer: 1) √2v m/s

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